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CHAPTER 1 PRELIMINARIES 



1.1 REAL NUMBERS AND THE REAL LINE 



1. Executing long division, ^ = 0.1, 



0.2, | = 0.3, | 



0.8, | 



0.9 



2. Executing long division, yj 



0.09, fy 



0.18, f\ 



27 — 



0.81, ii 



0.99 



NT = necessarily true, NNT = Not necessarily true. Given: 2 < x < 6. 

a) NNT. 5 is a counter example. 

b) NT. 2<x<6 ^2-2<x-2<6-2 =4>0<x-2<2. 

c) NT. 2 < x < 6 => 2/2 < x/2 < 6/2 => 1 < x < 3. 

d) NT. 2 < x < 6 => 1/2 > 1/x > 1/6 => 1/6 < 1/x < 1/2. 

e) NT. 2 < x < 6 => 1/2 > 1/x > 1/6 => 1/6 < 1/x < 1/2 -- 

f) NT. 2<x<6 => x<6=>(x-4)<2and2<x<6 
The pair of inequalities (x — 4) < 2 and — (x — 4) < 2 

g) NT. 2 < x < 6 => -2 > -x > -6 =4> -6 < -x < -2. But -2 < 2. So -6 < -x < -2 < 2 or -6 < -x < 2. 
h) NT. 2<x<6 => -1(2)>-I(x)<-1(6) => -6 < -x < -2 



6(1/6) <6(l/x)< 6(1/2) 

> x>2 => -x< -2 => 

> I x — 4 I < 2. 



> 1 < 6/x < 3. 

-x + 4<2 =4> 



(x-4)<2. 



4. NT = necessarily true, NNT = Not necessarily true. Given: — l<y — 5<1. 

a) NT. -1 <y - 5 < 1 =>-l+5<y-5 + 5<l+5 => 4 < y < 6. 

b) NNT. y = 5 is a counter example. (Actually, never true given that 4 < y < 6) 

c) NT. From a), — 1 < y - 5 < 1, =>4<y<6=^>y>4. 

d) NT. From a), - 1 < y - 5 < 1, =>4<y<6=>y<6. 

e) NT. — 1 <y — 5 < 1 =>-l + l<y-5+l<l + l =>0<y-4<2. 

f) NT. -I<y-5<1 => (l/2)(-l +5)<(l/2)(y- 5 + 5)< (l/2)(l+5) =>■ 2 < y/2 < 3. 

g) NT. From a), 4 < y < 6 =>■ 1/4 > 1/y > 1/6 => 1/6 < 1/y < 1/4. 

h) NT. — 1 < y — 5 < 1 => y - 5 > -1 => y>4 =>• -y < -4 => -y + 5 < 1 => -(y-5)<l. 

Also, — l<y — 5<1 => y — 5 < 1. The pair of inequalities — (y — 5) < 1 and (y — 5) < 1 =>• | y — 5 | < 1. 



5. -2x > 4 => x < -2 



6. 8 - 3x > 5 => -3x > -3 => x < 1 



-> x 



7. 5x - 3 < 7 - 3x => 8x < 10 => x < 



3(2 - x) > 2(3 + x) =>■ 6 - 3x > 6 + 2x 
=4> 0>5x => 0>x 



9. 2x-i>7x+|^>-i-|>5x 

=> l(-jf)>xor-i>i 



10. ^ < ^ => 12 - 2x < 12x - 16 

=4> 28 < 14x => 2 < x 




k)2 




2 Chapter 1 Preliminaries 

11. | (x - 2)< i (x - 6) =4> 12(x - 2) < 5(x - 6) 
=4> 12x - 24 < 5x - 30 =4> 7x < -6 or x < - | 



12. 



x±5 < i2±3x ^ _ (4x + 20) < 24 + 6x 

> -44 < lOx =*• - # < x 



-22/5 



13. y = 3 or y = — 3 



14. y - 3 = 7 or y - 3 = -7 =4> y = 10 or y = -4 

15. 2t + 5 = 4or2t + 5 = -4 => 2t=-lor2t=-9 =5> t=-| 



or t 



16. 1 - t = 1 or 1 - t = - 1 => -t = or -t = -2 =>■ t = or t = 2 



17. 8 - 3s = | or 8 - 3s = - § =4> -3s = - \ or -3s 



25 
2 



7 25 

S = 6 OT S = -6 



18. § - 1 = 1 or | - 1 = -1 => |=2 or |=0 => s = 4ors = 



19. —2 < x < 2; solution interval (—2, 2) 



20. —2 < x < 2; solution interval [—2, 2] 



-> x 



21. — 3 < t— 1 < 3 => -2 < t < 4; solution interval [-2, 4] 



22. — 1 < t + 2 < 1 => —3 < t < — 1; 

solution interval (—3, — 1) 



-»t 



23. -4 < 3y - 7 < 4 =^ 3 < 3y < 11 => 1 < y < ^ ; 



solution interval ( 1 , 



11/3 



24. -1 < 2y + 5 < 1 =>• -6 < 2y < -4 =4> 
solution interval (—3, —2) 



-3<y < -2; 



25. -1 < | - 1 < 1 =>• < f < 2 =^ < z < 10; 
solution interval [0, 10] 



26. -2 < f - 1 < 2 => -1 < f < 3 => 



3z 



| < z < 2; 



solution interval [— 1,2] 



-2/3 



27. 



I<3 



1 < i 

x ^ 2 



7 - < -1 < -5 

2x2 



7 15 

2 "^ x -^ 2 



| < x < | ; solution interval (1,1) 



28. -3 < ? - 4 < 3 =>• 1 < ? < 7 ^ 1 > | > i 



=4> 2 > x > | => 



2^7 

| < x < 2; solution interval (1,2) 



2/7 



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Section 1 . 1 Real Numbers and the Real Line 



29. 2s > 4 or -2s > 4 => s > 2 or s < -2; 
solution intervals (— oo, —2] U [2, oo) 



30. s + 3 > \ or -(s + 3) > \ => s > - \ or -s > \ 



=> s>— | or s < — | ; 
solution intervals 



-00,-^jU 



|,oo) 



-7/2 



-5/2 



31. 1 - x > 1 or -(1 - x) > 1 => -x > or x > 2 

=>• x<0orx>2; solution intervals (— oo, 0) U (2, oo) 



32. 2 - 3x > 5 or -(2 - 3x) > 5 => -3x > 3 or 3x > 7 
=4> x<— lorx>|; 
solution intervals (— oo, — 1) U (|, oo) 



7/3 



33. !±1 > 1 or 



'r+n 



> 1 =>• r+ 1 > 2orr+ 1 < -2 



=>• r > 1 or r < —3; solution intervals (— oo, —3] U [1, oo) 



34. |-1> |or-(f -1) > \ 

=^ ^>|or — ^>— | => r > | or r < 1 
solution intervals (— oo, 1) U ( |, oo) 



7/3 



35. x 2 < 2 => |x| < i/2 => -a/2 < x < a/2; 
solution interval I — a/2, a/ 2 ] 



-a/2 a/2 



36. 4 < x 2 => 2 < |x| =>• x > 2 or x < -2; 
solution interval (—00, —2] U [2, 00) 



-2 



37. 4<x 2 <9 =*> 2<|x|<3 => 2<x<3or2<-x<3 
=>■ 2 < x < 3 or -3 < x < -2; 
solution intervals (—3, —2) U (2, 3) 



38. i < x 2 < \ 



\<\x\<\ =4> 5<x<iorj<-x<i 



=>■ 5 < x < 1 or 
solution intervals (■ 



< x < 



5JU(f,f) 



1/2 -1/3 1/3 



1/2 



39. (x - l) 2 < 4 => |x - 1| < 2 => -2 < x - 1 < 2 
=>• — 1 < x < 3; solution interval (—1,3) 



40. (x + 3) 2 <2 => |x + 3| < a/2 

=>■ -a/2 <x + 3<\/2 or -3-a/2<x<-3 + a/2; 
solution interval I —3 — a/2, —3 + a/2 1 



-3 -a/2 -3+ a/2 



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Chapter 1 Preliminaries 



41. x 2 - x < => x — x+j< j => (x 
So the solution is the interval (0, 1) 



l) 2 <i 

2 J v 4 



1 < i 
2^2 



■\ <x - 1 < | => 0<x<l. 



42. x 2 - x - 2 > => x 2 



X+ 4-4^ X 2—2 



3 > | or -(x - i) > | =>• x > 2or x < -1. 



The solution interval is (— oo, — 1] U [2, oo) 

43. True if a > 0; False if a < 0. 

44. |x - 1| = 1 - x O |-(x - 1)| = 1 -x 44> 1 - x > <£► x < 1 

45. (1) |a + b| = (a + b) or |a + b| = -(a + b); 

both squared equal (a + b) 2 

(2) ab < |ab| = |a| |b| 

(3) |a| = a or |a| = —a, so |a| = a 2 ; likewise, |b| = b 2 

(4) x 2 < y 2 implies v x 2 < y> or x < y for all nonnegative real numbers x and y. Let x = |a + b| and 
y = |a| + |b| so that |a + b| 2 < (|a| + |b|) 2 =» |a + b| < |a| + |b| . 

46. If a > and b > 0, then ab > and |ab| = ab = |a| |b| . 

If a < and b < 0, then ab > and |ab| = ab = (-a)(-b) = |a| |b| . 
If a > and b < 0, then ab < and |ab| = -(ab) = (a)(-b) = |a| |b| . 
If a < and b > 0, then ab < and |ab| = -(ab) = (-a)(b) = |a| |b| . 

47. -3 < x < 3 and x > - \ => - \ < x < 3. 



48. Graph of |x| + |y| < 1 is the interior 
of "diamond-shaped" region. 



x| + |y|<l 




49. Let S be a real number > and f(x) = 2x + 1 . Suppose that | x — 1 \<6. Then | x— 1 | < 6 => 2| x— 1 | < 26 

| 2x - 2 | < 26 => | (2x + 1) - 3 | < 26 =4> | f(x) - f(l) | < 26 

50. Let e > be any positive number and f(x) = 2x + 3. Suppose that | x — | < e/2. Then 2| x — | < e and 
| 2x + 3 -3 | < e. But f(x) = 2x + 3 and f(0) = 3. Thus | f(x) - f(0) | < e. 



51. Consider: i) a > 0; ii) a < 0; iii) a = 0. 

i) For a > 0, | a | = a by definition. Now, a > => — a < 0. Let —a = b. By definition, | b | = — b. Since b = —a, 

| — a | = —(—a) = a and | a | = | — a | = a. 
ii) For a < 0, | a | = —a. Now, a < => — a > 0. Let —a = b. By definition, | b | = b and thus |— a| = —a. So again 

| a | = |— a|. 
iii) By definition | | = and since —0 = 0, | — | = 0. Thus, by i), ii), and iii) | a | = | — a | for any real number. 

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Section 1.2 Lines, Circles and Parabolas 



52. i) Prove | x | > =>■ x > a or x < — a for any positive number, a. 
For x > 0, | x | = x. | x | > a => x>a. 
Forx<0, |x| = — x. | x | > a =>■ — x > a => x < — a. 
ii) Prove x > a or x < — a =>• | x | > for any positive number, a. 
a > and x>a => |x| = x. So x > a =4> |x|>a. 
For a > 0, — a < and x < — a => x<0 => I x I = — x. So x < — a =4> — x > a =$■ I x I > a. 



53. a) 1 = 1 => | 1 

1 \ I ^ I 

b) nrr = a 



Mil 

b 



|b| 



54. Prove S n = |a n | = |a| n for any real number a and any positive integer n. 

la 1 ! = |a| : = a, so Si is true. Now, assume that Sk = |a k | = |a| k is true form some positive integer k. 

Since la 1 ! = lal : and |a k | = tat k , we have |a k+1 | = |a k ■ a 1 ! = |a k ||a 1 | = lal k lal x = |a| k+1 . Thus, 



S k +i = la 



k+ll 



a| is also true. Thus by the Principle of Mathematical Induction, S n = | a n | = | a | n 



is true for all n positive integers. 
1.2 LINES, CIRCLES, AND PARABOLAS 

1. Ax = -1 - (-3) = 2, Ay = -2 - 2 = -4; d = ^(Ax) 2 + (Ay) 2 = ^4+ 16 = 2 a/5 



2. Ax = -3 - (-1) = -2, Ay = 2 - (-2) = 4; d = ^/(-2) 2 + 4 2 = 2 a/5 

3. Ax = -8.1 - (-3.2) = -4.9, Ay = -2 - (-2) = 0; d = ^/(-4.9) 2 + 2 = 4.S 



4. Ax = 0- \pl= -s/l, Ay = 1.5-4 = -2.5; d 



5. Circle with center (0, 0) and radius 1. 



2) + (-2.5) 2 = a/^25 



6. Circle with center (0, 0) and radius y 2. 



7. Disk (i.e., circle together with its interior points) with center (0, 0) and radius y 3. 



8. The origin (a single point). 



9. m 



Ay 



-1-2 



Ax -2-(-l) 



10. m 



Ay 



-2-1 



Ax 2 -(-2) 



perpendicular slope 



perpendicular slope = | 








y 








4 










3 










2 








AN 


1 






-4 


-2 


-1 
-2 
-3 
-4 


2 


4 



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Chapter 1 Preliminaries 

Ay _ 3-3 _ q 

perpendicular slope does not exist 



«•»=£- -1-2 



12 - m =S = 3y^;no slope 
perpendicular slope = 



fi(-l, 3) 


A(2, 3) 


* 3 

2 


>' = 3 
Slope = 


1 


1 1 *x 





4 






3 








2 






B 


1 






-4 - 


2 
-1 


2 


4 


A - 2 








-3 








-4 







13. (a) x = -1 

(b) y = | 



14. (a) x = y/2 
(b) y = 1.3 



15. (a) x = 
(b) y=-y/2 



16. (a) x = — it 
(b) y = 



17. P(-l,l),m=-l => y-1 = -l(x-(-l)) => y = -x 



= ix-4 



18. P(2, -3), m = i => y - (-3) = \ (x - 2) => y = 5 x 

19. P(3,4),Q(-2,5) => m=^ = ^3 = -i => y_4 = -I(x-3) 

20. P(-8,0),Q(-1,3) => m = £* = ^^f^ = § => y- = § (x- (-8)) => y= fx+f 



' = -** 



I v J- 23 



21. m = -|,b = 6 => y = -| 
23. m = 0,P(-12,-9) => y = 



. + 6 



22. m=i,b=-3 =4> y=i x -3 
24. Noslope, P(|,4) =>• x= 1 

> y = 4x + 4 



i the line => m = ^ = -&=4 = 

Ax —1—0 



=>• y = -f x+ 1 



25. a=-l,b = 4 => (0,4) and (-1,0) are on i 

a = 2, b = -6 =>• (2, 0) and (0, -6) are on the line => m = ^ = z£^ =3 ^ y = 3x _ 6 
27. P(5,-l), L: 2x + 5y = 15 => m L = -| => parallel line is y - (-1) = -| (x - 5) => y 

= v / 3^m L = -^^ parallel line is y - 2 = -^ ( x _ (-1/2)) => y = 

P(4, 10), L: 6x - 3y = 5 => m L = 2 =4> m ± = -5 =>• perpendicular line is y - 10 = — | (x — 4) => y 
30. P(0, 1), L: 8x - 13y = 13 =4> m L = ^ => m ± = -^ =4> perpendicular line is y = -^ x + 1 



26 

28. P (-a/2,2) ,L: A/2x + 5y 
29. 



-^x + ^ 

5 S+ 5 



4x4- 12 



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31. x-intercept = 4, y-intercept = 3 



Section 1.2 Lines, Circles and Parabolas 
32. x-intercept = —4, y-intercept = —2 



t 


i 








V. 3*+4;y=12 






2 

1 











1 2 3 


is. 





33. x-intercept = y\3, y-intercept = — y 2 



t 


* . 







l/ 2 




-1 


"/ ~E.x-~&y = -fc 




-2 








34. x-intercept = —2, y-intercept = 3 





4 








/ 2 








/ 1 


1.5x-y. 


-3 


-4 , 


-1 
-2 
-3 

-4 


2 


4 



35. Ax + By = Ci 4* y = -§ x + §■ and Bx - Ay = C 2 O y= |x- &. Since (-§) (|) = -lis the 
product of the slopes, the lines are perpendicular. 

36. Ax + By = Ci <4> y = — ^ x + ^ and Ax + By = C 2 O y = — | x + ^. Since the lines have the same 



slope — § , they are parallel. 



37. New position = (x old + Ax, y ok , + Ay) = (-2 + 5,3 + (-6)) = (3, -3). 

38. New position = (x old + Ax, y„ kl + Ay) = (6 + (-6), + 0) = (0, 0). 

39. Ax = 5, Ay = 6, B(3, -3). Let A = (x, y). Then Ax = x 2 - x : => 5 = 3 - x => x = -2 and 
Ay = y 2 - yi => 6 = -3 - y => y = -9. Therefore, A = (-2, -9). 

40. Ax = 1 - 1 = 0, Ay = - = 



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8 Chapter 1 Preliminaries 

41. C(0, 2), a = 2 => x 2 + (y - 2) 2 = 4 



. 


J0.4) 


[ C(0,2)i 


h J 




(0,0)/ 


-2 -1 1 


' 1 2 * 



42. C(-3,0),a = 3 => (x + 3) 2 + y 2 




43. C(-l,5),a= V10 => (x + l) 2 + (y - 5) 2 = 10 




44. C(l, 1), a = 1/2 =>• (x - l) 2 + (y - l) 2 = 2 

x = => (0 - l) 2 + (y - l) 2 = 2 =>■ (y - l) 2 = 1 
=> y — 1 = ±1 => y = or y = 2. 
Similarly, y = =4> x = 0orx = 2 




45. c(- v / 3,-2),a = 2 => (x + ^3) + (y + 2) 2 = 4, 

x = => (0 + a/3~) + (y + 2) 2 = 4 => (y + 2) 2 = 1 
=> y + 2= ±1 =4> y = - 1 or y = -3. Also, y = 
=> U + a/3) + (0 + 2) 2 = 4 => (x + 1/3) = 
=> x= -a/3 

46. C (3, |), a = 5 =>• (x - 3) 2 + (y - i) 2 = 25, so 
x = =>• (0-3) 2 + (y- |) 2 = 25 

=> (y-i) 2 = 16 => y-|= ±4 => y=| 
ory = -|. Also, y = => (x - 3) 2 + (0 - \f = 25 
=> (x - 3) 2 = f => x - 3 = ± *-& 
=> x = 3±^- 



Copyright (c) 2006 Pearson Education, Inc 




(0,-3) 




Section 1.2 Lines, Circles and Parabolas 



47. x 2 + y 2 + 4x - 4y + 4 = 
=>■ x 2 + Ax + y 2 - 4y = -4 
=>■ x 2 + 4x + 4 + y 2 - 4y + 4 = 4 
=> (x + 2) 2 + (y - 2) 2 = 4 => C = (-2, 2), a 



frt + 2) 2 


+ G> 


y 

-2) 2 = 4 








-3 






• (0,2) -2 

C(-2, 2) 






(-2, o) y 




-4 


-i 


-2 -1 






48. x 2 + y 2 - 8x + 4y + 16 = 




=4> x 2 - 8x + y 2 + 4y = - 16 




=>• x 2 - 8x + 16 + y 2 + 4y + 4 = 


= 4 


=> (x - 4) 2 + (y + 2) 2 = 4 




=>• C = (4, -2), a = 2. 






49. x 2 + y 2 - 3y - 4 = =s> x 2 + y 2 - 3y 

=> x 2 + y 2 - 3y + § = f 

v2 



+ (y-|)' 



25 
4 



c=(o,|) 



a 




50. 


x 2 + y 2 


-4x-| = 


= 






=>■ 


x 2 - 


- 4x + y 2 = 


9 

4 






=>■ 


x 2 - 


- 4x + 4 + 


y 2 = 


25 
4 




=^ 


(x- 


- 2) 2 + y 2 -- 


25 
4 






=*> 


C = 


= (2,0),a = 


5 
2 - 






(4.5.0) 



51. x 2 + y 2 -4x + 4y = 

=>■ x 2 - 4x + y 2 + 4y = 

=4> x 2 - 4x + 4 + y 2 + 4y + 4 = 8 

=> (x - 2) 2 + (y + 2) 2 = 8 

=> C(2, -2), a = s/%. 



, 


* (x-2f + (y + 2f = $ 




(0,0) 


J^ \ (4,0) 




-1 / 
l-\ 


1 2 3 i\ 5 




-2 
V-3 


* 
C(2,-2) / 




-41 


(-4, 0) / 





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10 Chapter 1 Preliminaries 



52. x 2 + y 2 + 2x = 3 
=> x 2 + 2x + 1 - 



y 2 =4 
(x+l) 2 + y 2 = 4 
C = (-1,0), a = 2. 




53. x = -f 

2a 



-2 
2(1) 



1 



=> y = (l) 2 - 2(1) - 3 = -4 

=>. V = (l,-4). Ifx = 0theny = -3. 

Also, y = => x 2 -2x-3 = 
=>■ (x - 3)(x + 1) = =>■ x = 3or 

x = — 1. Axis of parabola is x — 1 . 




V(l,-4) 



54. x = - £ 



4 
2(1) 



=> y = (-2) 2 +4(-2) + 3 = -l 

^ V = (-2,-l). Ifx = 0theny = 3. 

Also, y = => x 2 +4x + 3 = 
=> (x + l)(x + 3) = => x = -1 or 

x = —3. Axis of parabola is x = —2. 




55 x - _ A - 4_ - ? 

JJ - A — 2a — 2(-l) ~~ z - 

=> y = -(2) 2 + 4(2) = 4 

=>■ V = (2,4). Ifx = 0theny = 0. 

Also, y = =^ -x 2 + 4x = 
=>■ -x(x - 4) = => x = 4orx = 0. 

Axis of parabola is x = 2. 



V(2, 4) 




56. 


* 2a 


2(-l) 






=► y = 


-(2) 2 +4(2)-5 = -l 






=>■ v = 


(2,-1). Ifx = 0theny = 






Also, y = 


= => -x 2 + 4x - 5 = 






=► x 2 - 


4x + 5 = =>• x = ^f 


4 



no x intercepts. Axis of parabola is x = 2. 




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Section 1.2 Lines, Circles and Parabolas 11 



57. x = - £ 



-3 



-6 

2(-l) 

(-3) 2 -6(-3)-5 = 4 



=>■ V = (-3,4). Ifx = Otheny = -5. 
Also, y = => -x 2 - 6x - 5 = 

=> (x + 5)(x + 1) = =4> x = -5 or 
x = — 1. Axis of parabola is x = —3. 




1(0,-5) 



58. x=- 



2a 



-1 

2(2) 



2Q) 2 -i + 3=f 



=> y 

=>. V= (i,f) . Ifx = 0theny = 3. 
Also, y = => 2x 2 -x + 3 = 



l±V-23 



no x intercepts. 



Axis of parabola is x 




59. x 



1 



2a 2(1/2) 

=> y = |(-i) 2 + (-i) + 4 = | 

=>■ V= (-1,|) ■ Ifx = 0theny = 4. 
Also, y = => |x 2 + x + 4 = 

=4> x = z — j^— =4- nox intercepts. 
Axis of parabola is x = — 1 . 



60. x = - £■ 

2a 



2(-l/4) _ ^ 

=> y = - 1 (4) 2 + 2(4) + 4 = 8 
=>. V = (4, 8) . If x = then y = 4. 
Also, y = => -jX 2 + 2x + 4 = 

=> *=^# =4 + 4^2. 
Axis of parabola is x = 4. 



(-2, 4) ** 
V(-l,7/2) 



(0,4) 
;y=^* 2 + j: + 4 



5> (4 + 4 >/2.o) 




61. The points that lie outside the circle with center (0, 0) and radius y 7. 

62. The points that lie inside the circle with center (0, 0) and radius y 5. 

63. The points that lie on or inside the circle with center (1, 0) and radius 2. 

64. The points lying on or outside the circle with center (0, 2) and radius 2. 

65. The points lying outside the circle with center (0, 0) and radius 1, but inside the circle with center (0, 0), 
and radius 2 (i.e., a washer). 



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12 Chapter 1 Preliminaries 



66. The points on or inside the circle centered 
at (0, 0) with radius 2 and on or inside the 
circle centered at (—2, 0) with radius 2. 




67. x 2 + y 2 + 6y < => x 2 + (y + 3) 2 < 9. 
The interior points of the circle centered at 
(0, —3) with radius 3, but above the line 

y = -3. 




68. x 2 + y 2 - 4x + 2y > 4 => (x - 2) 2 + (y + l) 2 > 
The points exterior to the circle centered at 
(2,-1) with radius 3 and to the right of the 
line x = 2. 




69. (x + 2) 2 + (y - l) 2 < 6 

71. x 2 + y 2 < 2, x > 1 

73. x 2 + y 2 = 1 and y = 2x => 1 = x 2 + 4x 2 = 5x 2 
=* (x = -*- andy = -^) or (x = --J- and y 

Thus ' A (7i'7;)' B (-7;'-7i) arethe 

points of intersection. 



70. (x + 4) 2 + (y - 2) 2 > 16 

72. x 2 + y 2 > 4, (x - l) 2 + (y - 3) 2 < 10 



*) 



2 2 




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Section 1.2 Lines, Circles, and Parabolas 13 



74. x + y = 1 and (x - l) 2 + y 2 = 
=> 1 = (-y) 2 + y 2 = 2y 2 

4= and x = 1 ! 



y "75 """ A -^75. 

(y = -^ andx=l + 4^). Thus, 

A ( 1 -75'73) andB ( 1 + 73'-7i) 

are intersection points. 



( x -1) 2 H.y 2 = l 




75. y — x = 1 and y = x =>• x — x = 1 
=> x 2 -x-l=0 => x = ^ . 



Ifx 



1 + %/s 



then y = x + 1 



If x = ±^ , then y = x + 1 



3+y/5 
2 

3-n/5 



Thus 



,A( 



1+V5 3+^ andB^-VS 3-V5 



2 > 2 

are the intersection points 




76. y = — x and y = — (x — l) 2 =>- (x — l) 2 



=>■ x 2 - 3x + 1 







x = ^ . If 



2 
3+A/5 



then y 
then y 



/5-3 



If 



3+V5 



Thus,A(^ J 4-)andB(^,-^) 



2 ' 2 

are the intersection points. 




77. y = 2x 2 - 1 = -x 2 => 3x 
=> x 



4- and y = — i or x = \- and y = — I 

Vi J 3 V3 J 3 

Thus, A (4- , - 1 J and B (- 4; , - i J are the 
intersection points. 




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14 Chapter 1 Preliminaries 



78. y 



(x - l) 2 







3x 2 



-2x+ 1 



= 3x 



2 



4 = (3x - 2)(x - 2) 



1 , or x = | and 



x = 2 and y = \ 

% = \. Thus, A(2,l) and B(§,± 



are the intersection points. 




79. x 2 + y 2 = 1 = (x - l) 2 - 

=>. X 2 = (x- l) 2 =x 2 - 
=> = -2x + 1 => x 



y 

2x+ 1 
= 1 Hence 



1 



| ory 



± 



/3 



Thus, 



A(±,^) mxdBU,-^) are the 
intersection points. 



A (x-l) 2 +y 2 -l 





80. x 2 -J- " 2 



x 2 + y => y 2 



y 



y = or y 

-y 2 



i. 



r = i 

=> y(y - 1) = 

If y = 1, then x 2 = 1 — y 2 = or x = 0. 
If y = 0, then x 2 = 1 — y 2 = 1 or x = ± 1. 
Thus, A(0, 1), B(l, 0), and C(-l, 0) are the 
intersection points. 




1 -x' 



$1. (a) A«(69°,0in), B w (68°, .4 in) 

(b) A w (68°, .4 in), B « (10°, 4 in) 

(c) A w (10°, 4 in), B w (5°, 4.6 in) 



68° - 69° 



m 



.4 


-0 


10° 


-68° 


4- 


-.4 


5°- 


10° 



4.6-4 



-2.5°/in. 
-16.17in. 

^8.3°/in. 



82. The time rate of heat transfer across a material, -^ , is directly proportional to the cross-sectional area, A, of the material, 
to the temperature gradient across the material, ^ (the slopes from the previous problem), and to a constant characteristic 

of the material. 4j = -kA^J =4- k = — Hj- . Note that 4j and ^ are of opposite sign because heat flow is toward lower 

temperature. So a small value of k corresponds to low heat flow through the material and thus the material is a good 
insulator.Since all three materials have the same cross section and the heat flow across each is the same (temperatures are 
not changing), we may define another constant, K, characteristics of the material: K = — -^ . Using the values of ^ from 

At 

the prevous problem, fiberglass has the smallest K at 0.06 and thus is the best insulator. Likewise, the wallboard is the 
poorest insulator, with K = 0.4. 



83. p = kd + 1 and p = 10.94 at d = 100 
pressure equation so that d = 50 



t _ 10.94-1 
K 100 

p = (0.0994)(50) 



0.0994. Then p = 0.0994d + 1 is the diver's 
1 = 5.97 atmospheres. 



84. The line of incidence passes through (0, 1) and (1,0) => The line of reflection passes through (1, 0) and (2, 1) 
=$■ m = ^5j = 1 =** y — = l(x — 1) =^> y = x— lis the line of reflection. 



Copyright (c) 2006 Pearson Education 





160 



160 
9 



Section 1.2 Lines, Circles, and Parabolas 15 
or F = —40° gives the same numerical reading. 



86. m 



37.1 



_14 
Ax 



Ax = ~j. Therefore, distance between first and last rows is a/(14) 2 + (4lr) « 40.25 ft. 



87. length AB = a/(5 - l) 2 + (5 - 2) 2 = a/T6 
length AC = a/(4 - l) 2 + (-2 - 2) 2 
length BC = a/(4 - 5) 2 + (-2 - 5) 2 



9 = 5 



a/9 +16 = 5 
V / TT49= a/50 



5V2^5 



88. length AB = J(l - 0) 2 + (a/3 - o) = a/1+3 = 2 
length AC = a/(2 - 0) 2 + (0 - 0) 2 = a/4 + = 2 
length BC 



\y(2-D 2 +(0-A/3) 2 = A/ 



1+3 



a/i 2 + 4 2 = a/17 and length BC = a/(Ax) 2 + (Ay) 2 = a/4 2 + l 2 



89. Length AB = a/(Ax) 2 + (Ay) 2 = a/ I 2 + 4 2 = a/ 17 and length BC = a/(Ax) 2 + (Ay) 2 = a/4 2 + l 2 = a/ 17. 
Also, slope AB = ^y and slope BC = 4, so AB ± BC. Thus, the points are vertices of a square. The coordinate 
increments from the fourth vertex D(x, y) to A must equal the increments from CtoB => 2 — x = Ax = 4 and 
— 1 — y = Ay = 1 =>■ x = —2 and y = —2. Thus D(— 2, —2) is the fourth vertex. 



90. Let A = (x, 2) and C = (9, y) => B = (x, y). Then 9 - x = |AD| and 2 - y = |DC| => 2(9 - x) + 2(2 - y) = 56 
and 9 - x = 3(2 - y) =* 2(3(2 - y)) + 2(2 - y) = 56 => y = -5 => 9 - x = 3(2 - (-5)) =4> x = -12. 
Therefore, A = (-12, 2), C = (9, -5), and B = (-12, -5). 



91. Let A(-l, 1), B(2, 3), and C(2, 0) denote the points. 
Since BC is vertical and has length |BC| = 3, let 
Di(— 1,4) be located vertically upward from A and 
D2(— 1, —2) be located vertically downward from A so 
that |BC| = |ADi| = |AD 2 | = 3. Denote the point 
Da(x, y). Since the slope of AB equals the slope of 
CD 3 we have £E§ = ~ 3 => 3y-9 = -x + 2or 
x + 3y = 11. Likewise, the slope of AC equals the slope 
of BD 3 so that |f| = § => 3y = 2x - 4 or 2x - 3y = 4. 

x + 3y 



Solving the system of equations 



2x-3y 



"} 



(-1,-2)* -2 



• (5,2) 




we find x = 5 and y = 2 yielding the vertex D3(5, 2). 



92. Let (x,y),x^0 and/or y 7^ be a point on the coordinate plane. The slope, m, of the segment (0, 0) to (x, y) is - . A 90° 
rotation gives a segment with slope m' = — j- = — | . If this segment has length equal to the original segment, its endpoint 
will be (— y, x) or (y, — x), the first of these corresponds to a counter-clockwise rotation, the latter to a clockwise 
rotation. 

(a) (-1,4); (b) (3,-2); (c) (5,2); (d) (0,x); 

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16 Chapter 1 Preliminaries 

(e) (-y,0); (f) (-y,x); (g) (3,-10) 



93. 2x + ky = 3 has slope — | and 4x + y = 1 has slope —4. The lines are perpendicular when — | (—4) 



1 or 



k = — 8 and parallel when 



-4 or k 



2' 



94. At the point of intersection, 2x + 4y = 6 and 2x — 3y = — 1 . Subtracting these equations we find 7y = 7 or 
y = 1. Substitution into either equation gives x = 1 =$■ (1,1) is the intersection point. The line through (1, 1) 
and (1, 2) is vertical with equation x = 1. 



95. Let M(a, b) be the midpoint. Since the two triangles 
shown in the figure are congruent, the value a must 
lie midway between Xi and X2, so a 



*l+*2 

2 ■ 



Similarly, b= 2d£>. 



....P^.y,) 



M(a,b) 




Q(x 2 ,y 2 ) 



96. (a) L has slope 1 so M is the line through P(2, 1) with slope —1; or the line y = 
point, Q, we have equal y-values, y = x + 2 = — x + 3. Thus, 2x = 1 or x 



'I 1) 

,2> 2) 



The distance from P to L = the distance from P to Q 



x + 3. At the intersection 
I . Hence Q has coordinates 

. V2 



i)' 



(b) L has slope — | so M has slope | and M has the equation 4y — 3x = 12. We can rewrite the equations of 



the lines as L: x + | y = 3 and M 
into either equation gives x 



-x+ fy 



4. Adding these we get y| y = 7 so y 



|| . Substitution 



4 (S4\ 

3 I 25/ 



4 = || so that Q (si' If) * s me P omt of intersection. The distance 
fromPtoL=^(4-i|) 2 + (6-l) 2 = f. 

(c) M is a horizontal line with equation y = b. The intersection point of L and M is Q(— 1, b). Thus, the 
distance from P to L is ^/(a + l) 2 + 2 = |a + 1 1 . 

(d) If B = and A^O, then the distance from P to L is | j - x 1 as in (c). Similarly, if A = and B^O, the 

A 



distance is I £ 



I A 

yo | . If both A and B are ^ then L has slope -gSoM has slope 



|. Thus, 



L: Ax + By = C and M: — Bx + Ay = — Bxo + Ayo. Solving these equations simultaneously we find the 



point of intersection Q(x, y) with x 



AC-B (Ayp-Bxp) 



A 2 +B 2 



and y 



BC+A(Ay -Bx ) 
A 2 +B 2 



The distance from 



P to Q equals ^(Ax) 2 + (Ay) 2 , where (Ax) 2 = ^ (A 2 +B 2 )-AC+AB yo -B% y 



A 2 (Ax +By +C) : 
(A 2 +B 2 ) 2 



, and (Ay) 2 = ( 



yo (A 2 +B 2 )-BC-A 2 yo+ABx ' 
A 2 +B 2 



B 2 (Axp+Byo+C) 2 
(A 2 +B 2 ) 2 



Thus, v/(Ax) 2 + (Ay) 2 



(Axp+Byp+C) 2 
A 2 +B 2 



|Ax +Byo+C| 

v'a 2 +b 2 



1.3 FUNCTIONS AND THEIR GRAPHS 



1. domain = (—oo, oo); range = [1, oo) 



2. domain = [0, oo); range = (— oo, 1] 



3. domain = (0, oo); y in range => y=^ r ,t>0 =^ y 2 = \ and y > =^ y can be any positive real number 
=^ range = (0, oo). 



Copyright (c) 2006 Pearson Education 




4. domain = [0, oo); y in range 

and smaller positive real number =>■ range = (0, 1] 



Section 1.3 Functions and Their Graphs 17 

y = — A- , t > 0. If t = 0, then y = 1 and as t increases, y becomes a smaller 

J i+\A 



5. 4 — z 2 = (2 — z)(2 + z) > <^ z 6 [—2, 2] = domain. Largest value is g(0) = y 4 = 2 and smallest value is 
g(-2) = g(2) = v/0 = =► range = [0, 2]. 

6. domain = (—2, 2) from Exercise 5; smallest value is g(0) = \ and as < z increases to 2, g(z) gets larger and 
larger (also true as z < decreases to —2) =4> range = [|, oo) . 

7. (a) Not the graph of a function of x since it fails the vertical line test. 

(b) Is the graph of a function of x since any vertical line intersects the graph at most once. 

8. (a) Not the graph of a function of x since it fails the vertical line test, 
(b) Not the graph of a function of x since it fails the vertical line test. 



9. y 



1 => - - 1 > =>• x < 1 and x > 0. So, 



(a) No (x > 0); (b) No; division by undefined; 

(c) No; if x > 1, i < 1 => i - 1 < 0; (d) (0, 1] 



10. y = a/2 - a/x => 2 - a/x > => a/x > and a/x < 2. a/x > => x > and a/x < 2 =4> x < 4. So, < x < 4. 
(a) No; (b) No; (c) [0, 4] 



11. base = x; (height) 2 + (f) = x 2 => height = \- x; area is a 
perimeter is p(x) = x + x + x = 3x. 



(x) = 1 (base)(height) = \ (x) (^x) = ^ x 2 ; 



12. s = side length => s 2 + s 2 = d 2 =^ s = A- ; and area is a = s 2 =^> a = | d 2 



13. Let D = diagonal of a face of the cube and t = the length of an edge. Then I + D = d" and (by Exercise 10) 



D 2 = 2l 2 => 3f = d 2 ^ £ = A ■ The surface area is 6f = ^# = 2d 2 and the volume 



is£ 3 =(f) 



3/2 



d 3 



14. The coordinates of P are (x, ^x) so the slope of the line joining P to the origin is m = ^ = A- (x > 0). Thus, 




16. The domain is (— oo, oo). 





f( 


x) 




3 




f(x) = 


l-2x-x 2 






X~ "N, 2 






/ - 1 




-3 / 


-2 -1 


\ 1 




-1 






-2 





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18 Chapter 1 Preliminaries 
17. The domain is ( — oo, oo). 




19. The domain is ( — oo, 0) U (0, oo). 



F(l) = 



12 3 4 



18. The domain is (— oo, 0]. 




20. The domain is (— oo, 0) U (0, oo). 

f(x) 

1 




21. Neither graph passes the vertical line test 
(a) 




22. Neither graph passes the vertical line test 
(a) 




(b) 




(b) 




f x + y = i | [y=i-x| 

< or > <^> < or > 



x + y 
|x + y| = 1 <^> I or 

[ x + y =_lj ^y= l xj 



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Section 1.3 Functions and Their Graphs 19 



23. 



X 





1 


2 


y 





1 







24. 



X 





i 


2 


y 


i 










r 1 -x , o<x<" 

l 2-x,1 <x<2 



25. y 



3 - x, x < 1 
2x, 1 < x 








2 
1 


. F(x) = 


3 - x, x<l 

2x, x>l 




-3 -2 -1 


1 2 3 


• (a) 


Line through (0, 0) and (1, 1): y = x 




Line through (1, 1) and (2. 0): y = -x + 2 




fl , , f x, < x < 1 
W -\-x + 2, 1 < x < 2 






'2, < x < 1 


(b) 


f(x) = < 


0, 1 < x < 2 
2, 2 < x < 3 






0, 3 < x < 4 







26. y 



\> x<0 
x, < x 




28. (a) Line through (0, 2) and (2, 0): y 
Line through (2, 1) and (5, 0): m 

J -x 4- 2, < x < 2 
f(x)= \ -ix+jj, 2<x<5 
(b) Line through (-1, 0) and (0, -3): m 
Line through (0, 3) and (2, —1): m = 
' -3x-3, -1 < x < 



o- 1 

5-2 



3 



-3-0 
" 0-(-l) 

-1-3 
2-0 



f(x) 



-2x- 



< x < 2 



so y 



|(x-2) + l 



-3, so y = — 3x — 3 

= -2, soy = -2x4-3 



29. (a) Line through (-1, 1) and (0, 0): y = -x 
Line through (0, 1) and (1, 1): y = 1 
Line through (1, 1) and (3, 0): m = §=i 



2 



so y 



•i(x- 1)4-1 



—x 

1 



f(x) = 

I 2^ ' 2 

(b) Line through (—2, 



-1 < x < 
< x < 1 
1< x < 3 
-l)and(0, 0):y 



oX 



Line through (0, 2) and (1, 0): y = -2x 4- 2 
Line through (1, —1) and (3, —1): y = — 1 



- „x + „ 



Copyright (c) 2006 Pearson Education 




20 Chapter 1 Preliminaries 



f(x) 



Iv 

2 x 



-2 < x < 
-2x + 2 < x < 1 
^ -1 1 < x < 3 



30. (a) Line through (f , 0) and (T, 1): m 



1-0 
T-(T/2) 



so y 



f(x) 



(b) f(x) = i 



0, < x < | 
ix-1, I < x < T 

A, < x < | 



-A, 



< x < T 

3T 



A, T < x < 

-A, f < x < 2T 



• = ix-i 



31. (a) From the graph, | > 1 



xe (-2,0)U(4,oo) 



(b) | > 1 



=> 5-1-4 >Q 



2 

x>0: | - 1 - 1 > ^ ^f=^ > 
=^> x > 4 since x is positive; 

4 >Q => ^!4^8 <0 



x < 0: f - 1 



2x 



x < —2 since x is negative; 
sign of (x - 4)(x + 2) 



^t— 



-2 4 

Solution interval: (-2, 0) U (4, oo) 



(x-4)(x+2) n 

2x ^ U 

(x-4)(x+2) n 

2x v u 




g{x) = \*1 



f(x) -\ 



32. (a) From the graph, ^y < ^y =4> x e (-oo, -5) U (-1, 1) 
(b) Casex<-1: £ < £ => ^ > 2 

=>• 3x + 3 < 2x - 2 => x < -5. 
Thus, x G (— oo, —5) solves the inequality. 



Case - 1 < x < 1: r^r < 2 



3(i+l) 



<2 



x-l x+1 x-1 

=4> 3x + 3 > 2x — 2 =4> x > — 5 which is true 

if x > — 1. Thus, x G (—1,1) solves the 

inequality. 
Case l<x: ^t < ^y => 3x + 3<2x-2 => x<-5 

which is never true if 1 < x, so no solution 

here. 
In conclusion, x e (— oo, —5) U (— 1, 1). 




33. (a) [xj =0forxe [0,1) 



(b) |Y| =0forxe (-1,0] 



34. L X J = |~ x l only when x is an integer. 

35. For any real number x, n < x < n + 1, where n is an integer. Now: n<x<n+l=>— (n + 1) < — x < — n. By 
definition: |~— x] = — n and |_xj = n => — |_xj = — n. So |~— x] = — |_xj for all x e 5R. 



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Section 1.3 Functions and Their Graphs 21 



36. To find f(x) you delete the decimal or 
fractional portion of x, leaving only 
the integer part. 



-3- 



-i 

-2| f(x}. 
-3 



|xj, x>.0 

r*T > x < ° 



37. v = f(x) = x(14 - 2x)(22 - 2x) = 4x 3 - 72x 2 + 308x; < x < 7. 



38. (a) Let h = height of the triangle. Since the triangle is isosceles, AB 2 + AB ' 



AB 



'2. So, 



h 2 + l 2 = (y/2) => h = 1 => B is at (0, 1) =>■ slope of AB = -1 => The equation of AB is 

y = f(x) = -i+l;xe [0, 1]. 
(b) A(x) = 2xy = 2x(-x + 1) = -2x 2 + 2x; x E [0, 1]. 



39. (a) Because the circumference of the original circle was 8tt and a piece of length x was removed. 

X 

2rr 



(b)r=^=4-i 

(c) h = v/16^7 2 " = 7l6-(4-^) 2 = ^16- (16- £+*,)-. A* - ^i - - /il- - --- ^^ 

(d) V=^r 2 h=l7r( 2 



4ir 2 V 47r 2 47r 2 



i ,,2 1, : „( 8ir-x \ 2 \/l67rx - x 2 _ (87r - x) 2 Vl67rx - x 2 

2?r — 24?r 2 



40. (a) Note that 2 mi = 10,560 ft, so there are V800 2 + x 2 feet of river cable at $180 per foot and (10, 560 - x) feet of land 
cable at $100 per foot. The cost is C(x) = 180 v / 800 2 ~+x 2 + 100(10, 560 - x). 
(b) C(0) = $1,200,000 
C(500) « $1,175, 812 
C(1000) « $1,186, 512 
C(1500) « $1,212,000 
C(2000) w $1,243, 732 
C(2500) « $1,278,479 
C(3000) w $1,314,870 

Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the 
point P. 



41. A curve symmetric about the x-axis will not pass the vertical line test because the points (x, y) and (x, — y) lie on the same 
vertical line. The graph of the function y = f(x) = is the x-axis, a horizontal line for which there is a single y-value, 0, 
for any x. 



42. Pick 1 1, for example: 11 + 5 = 16 -» 2 • 16 = 32 ->■ 32 - 6 = 26 



f(x) 



2(x+5)-6 



2 = x, the number you started with. 



Y = 13— >13 — 2 = 11, the original number. 



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22 Chapter 1 Preliminaries 

1.4 IDENTIFYING FUNCTIONS; MATHEMATICAL MODELS 



1. (a) linear, polynomial of degree 1, algebraic. 
(c) rational, algebraic. 



(b) power, algebraic, 
(d) exponential. 



2. (a) polynomial of degree 4, algebraic, 
(c) algebraic. 



(b) exponential. 

(d) power, algebraic. 



3. (a) rational, algebraic, 
(c) trigonometric. 



(b) algebraic, 
(d) logarithmic. 



4. (a) logarithmic. 
(c) exponential. 



(b) algebraic, 
(d) trigonometric. 



5. (a) Graph h because it is an even function and rises less rapidly than does Graph g. 

(b) Graph f because it is an odd function. 

(c) Graph g because it is an even function and rises more rapidly than does Graph h. 

6. (a) Graph f because it is linear. 

(b) Graph g because it contains (0, 1). 

(c) Graph h because it is a nonlinear odd function. 



7. Symmetric about the origin 
Dec: — oo < x < oo 
Inc: nowhere 



Symmetric about the y-axis 
Dec: — oo < x < 
Inc: < x < oo 





9. Symmetric about the origin 
Dec: nowhere 
Inc: — oo < x < 
< x < oo 



10. Symmetric about the y-axis 
Dec: < x < oo 
Inc: — oo < x < 



-2-1 



-2 



-li 




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Section 1.4 Identifying Functions; Mathematical Models 23 



11. Symmetric about the y-axis 
Dec: — oo < x < 
Inc: < x < oo 



12. No symmetry 

Dec: — oo < x < 
Inc: nowhere 




■vn 



13. Symmetric about the origin 
Dec: nowhere 
Inc: — oo < x < oo 



14. No symmetry 
Dec: < x < oo 
Inc: nowhere 





t 


. 




1 


~y- 




i 1/8 


_ 


2 / 


^■""-1/8 




/ 


-1 


■ 





15. No symmetry 
Dec: < x < oo 
Inc: nowhere 



y 








. 


, 











\ 


2 


3 




-l 

-2 




\ y ^ 


3/2 
-X 




-3 










-4 










-5 











16. No symmetry 




Dec: — oo < x < 




Inc: nowhere 






y 




25 




20 




15 


3/2 X. 

y-H> N. 


10 




5 


-8 -6 -4 


-2 




-5 




-10 



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24 Chapter 1 Preliminaries 



17. Symmetric about the y-axis 
Dec: — oo < x < 
Inc: < x < oo 




18. Symmetric about the y-axis 
Dec: < x < oo 
Inc: — oo < x < 





y 








10 










7.5 










s 










2.5 








-20 


-io ; 
-2/5 

s " 5 




10 


20 




-7.5 


t~ 


2/3 
-X 






-10 









19. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the 
function is even. 



20. f(x) = x- 5 = \ and f(-x) = (-x) 



-5 _ 1 

(-x) 5 



'^\ = — f(x). Thus the function is odd. 



21. Since f(x) = x 2 + 1 = (— x) + 1 = — f(x). The function is even. 

22. Since [f(x) = x 2 + x] ^ [f(— x) = (— x) — x] and [f(x) = x 2 + x] 7^ [— f(x) = — (x) — x] the function is neither even nor 
odd. 

23. Since g(x) = x 3 + x, g(— x) = — x 3 — x = — (x 3 + x) = — g(x). So the function is odd. 

24. g(x) = x 4 + 3x 2 + 1 = (— x) + 3(— x) — 1 = g(— x), thus the function is even. 

25. g(x) = ^rzrj = ,_ I2, = g( — x )- Thus the function is even. 

26. g(x) = ^ri; g(-x) = - j^ = g(-x). So the function is odd. 

27. h(t) = ^j; h(-t) = ~^zrj; -h(t) = j-L. Since h(t) ^ -h(t) and h(t) ^ h(-t), the function is neither even nor odd. 

28. Since 1 1 3 | = | (-t) 3 |, h(t) = h(-t) and the function is even. 

29. h(t) = 2t + 1, h(-t) = -2t + 1. So h(t) ^ h(-t). -h(t) = -2t - 1, so h(t) ^ -h(t). The function is neither even nor 
odd. 

30. h(t) = 2| 1 1 + 1 and h(-t) = 2| -t | + 1 = 2| 1 1 + 1. So h(t) = h(-t) and the function is even. 



31. (a) 



y = 0.166.t 



The graph supports the assumption that y is proportional to x. The 
constant of proportionality is estimated from the slope of the 
regression line, which is 0.166. 




Section 1.4 Identifying Functions; Mathematical Models 25 



(b) 



The graph supports the assumption that y is proportional to x 1 / 2 . 
The constant of proportionality is estimated from the slope of the 
regression line, which is 2.03. 



32. (a) Because of the wide range of values of the data, two graphs are needed to observe all of the points in relation to the 
regression line. 

y 




500 1000 1500 2000 2500" 




The graphs support the assumption that y is proportional to 3 X . The constant of proportionality is estimated from the 
slope of the regression line, which is 5.00. 
(b) The graph supports the assumption that y is proportional to In x. The constant of proportionality is extimated from 
the slope of the regression line, which is 2.99. 



20 

15 

10 

5 



In x 



33. (a) The scatterplot of y = reaction distance versus x = speed is 




10 20 30 40 50 60 70 80 90100 

Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is 
approximately 1.1. 



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26 Chapter 1 Preliminaries 

(b) Calculate x' = speed squared. The scatterplot of x'versus y = braking distance is: 



400 




6400 



Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which 
is approximately 0.059. 

34. Kepler's 3rd Law is T(days) = 0.41R 3 ' 2 , R in millions of miles. "Quaoar" is 4 x 10 9 miles from Earth, or about 
4 x 10 9 + 93 x 10 6 « 4 x 10 9 miles from the sun. Let R = 4000 (millions of miles) and 
T = (0.41)(4000) 3/2 days w 103, 723 days. 



35. (a) 



y 
















■ 


















10 


















9 
















• 


8 
















• 


7 
















• 


6 
















• 


5 














• 




4 
3 








• 


m 


• 






2 
1 




• 


• 

















1 


2 


1 


4 


5 


6 


7 8 9 10 



8.741 -0 



The hypothesis is reasonable. 

(b) The constant of proportionality is the slope of the line ~ 

(c) y(in.) = (0. 87 in./unit mass)(13 unit mass) = 11.31 in. 



in./unit mass = 0.874 in./unit mass. 



36. (a) 



(b) 



300 



200 



100 



300 



200 



100 



• • 



500 1000 1500 



20000 40000 60000 



Graph (b) suggests that y = kx 3 is the better model. This graph is more linear than is graph (a). 
1.5 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS 

1. D, : -oo < x < oo, D g : x > 1 => D r+g = D lg : x > 1. R, : -oo < y < oo, R g : y > 0, R 1+g : y > 1, R lg : y > 

2. D, : x + 1 > => x > -1, D g : x - 1 > => x > 1. Therefore D f+g = D fg : x > 1. 
Rf = R g : y > 0, R f+g : y > y/l, R lg : y > 



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Section 1.5 Combining Functions; Shifting and Scaling Graphs 27 

3. D f : — oo < x < oo, D g : — oo < x < oo =>• D,y g : — oo < x < oo since g(x) ^ for any x; D g/ , : — oo < x < oo 
since f(x) / for any x. R, : y = 2, R„: y > 1, R f/g : < y < 2, R g/f : y > \ 

4. D f : -oo < x < oo, D g : x > =4> D f/g : x > since g(x) ^ for any x > 0; D g/f : x > since f(x) ^ 
for any x > 0. R f : y = 1, R E : y > 1, R l/g : < y < 1, R g/f : y > 1 



5. (a 
(b 

(c 

(d 

(c 

(f) 

( 

(h 

6. (a 

(b 
(c 

(d 
(c 
(f) 

( 

(h 

7. (a 
(b 

(c 
(d 
(e 
(f) 

8. (a 

(b 
(c 
(d 
(c 
(f) 

9. (a 

(c 

(c 

10. (a 

(c 

(c 



f(g(0)) = f(-3) = 2 

g(f(0)) = g(5) = 22 

f(g(x)) = f(x 2 - 3) = x 2 - 3 + 5 = x 2 + 2 

g(f(x)) = g(x + 5) = (x + 5) 2 - 3 = x 2 + lOx + 22 

f(f(-5)) = f(0) = 5 

g(g(2)) = g(l) = -2 

f(f(x)) = f(x + 5) = (x + 5) + 5 = x + 10 

g(g(x)) = g(x 2 - 3) = (x 2 - 3) 2 - 3 = x 4 - 6x 2 + 6 



f(g(i» 

g(f(D) 



'(f) 



{-\) 



ffe(x))=f(ji T ): 

g(f(x)) = g(x - 1) 
f(f(2)) = f(l) = 
g(g(2)) 



l 

x+I 



x+1 

1 



(x-l)+l 



8 (J) 



f(f(x)) = f(x - 1) = (X - 1) - 1 

g(g(x)) = g (rrr) = 



-+1 



x+1 

x + 2 



(x/ -1 andx^ -2) 



5- ± 



u(v(f(x))) = u(v(I))=u(i)=4(i) 
u(f(v(x))) = u (f (x 2 )) = u (i) = 4 (i) - 5 
v(u(f(x))) = v(u(I))=v(4(i)-5) = ( 4 -5) 
v(f(u(x))) = v(f(4x - 5)) = v (3^5) 
f(u(v(x))) = f(u(x 2 ))=f(4(x 2 )-5 
f(v(u(x))) = f(v(4x - 5)) = f ((4x - 5) : 



.x 

lix^li 

1 

4x 2 -5 
\2\ _ 1 



2 



(4x - 5) 2 



h(g(f(x))) = h (g (v^)) = h (^) = 4 (^) 
h(f(g(x))) = h (f (I)) = h (v% = VJ - 8 
g(h(f(x))) = g (h (y^)) = g (4^/x" - 8) 
g(f(h(x))) = g(f(4x - 8)) = g (^4x-8) 
f(g(h(x))) = f(g(4x - 8)) = f (^) = f(x - 2) 
f(h(g(x))) = f(h(|))=f(4(|)-8) 

y = f(g(x)) 

y = g(g(x)) 

y = g(h(f(x))) 

y = fQ(x)) 
y = h(h(x)) 

y=j(g(f(x))) 



8 = 2^ 



ygx - X 
4 



'x-2 

y/x-2 

2 



f(x - 8) = Vx" 



(b) y=j(g(x)) 

(d) y = jQ(x)) 

(f) y = hG(f(x))) 

(b) y = h(g(x)) = g(h(x)) 

(d) y = f(f(x)) 

(f) y = g(f(h(x))) 



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28 Chapter 1 Preliminaries 

11. g(x) f(x) 



(f°g)(x) 



(a) x - 7 



(b) 


x + 2 


(c) 


x 2 


(d) 


X 

x-1 


(c) 


1 

x-1 


(f) 


1 



3x 

X 
X- 1 

1 + i 

X 

1 



3(x + 2) = 3x + 6 

A 2 -5 



-1 x-(x-l) 



X 
X 



12. (a) (fog)(x) = |g(x)| = ^. 

(b) (f°g)(x) = ^ = ^^1-^ = ^^!-^ = ^^ = ^sog(x) = x+ 1. 

(c) Since (fog)(x) = Vg(x) = |x|, g(x) = x 2 . 

(d) Since (fog)(x) = f(\/x) = |x |, f(x) = x 2 . (Note that the domain of the composite is [0, oo).) 
The completed table is shown. Note that the absolute value sign in part (d) is optional. 



g(x) 


f(x) 


(f°g)(x) 


1 

x-1 


W 


1 

|x-l| 


X+l 


x-1 

X 


X 
X+l 


X 2 


^ 


|x| 


V^ 


X 2 


|x| 



13. (a) f(g(x)) = ^[TT 
g(f(x)) 



l 

/x+l 

(b) Domain (fog): (0, oo), domain (gof): (— 1, oo) 

(c) Range (fog): (1, oo), range (gof): (0, oo) 

14. (a) f(g(x)) = 1-2-y/x + x 

g(f(x)) = 1 - |x| 

(b) Domain (fog): (0, oo), domain (gof): (0, oo) 

(c) Range (fog): (0, oo), range (gof): (— oo, 1) 

15. (a) y = -(x + 7) 2 

16. (a) y = x 2 + 3 



(b) y = -(x - 4) 2 
(b) y = x 2 - 5 



17. (a) Position 4 (b) Position 1 (c) Position 2 (d) Position 3 

18. (a) y = -(x-l) 2 + 4 (b) y = -(x + 2) 2 + 3 (c) y=-(x + 4) 2 -l (d) y = -(x - 2) 2 



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Section 1.5 Combining Functions; Shifting and Scaling Graphs 29 



19. 



20. 




(* + 4)'t(y-3) , -25 y 



\ ^*- "^^ B 




I / 

\ 1 


ii +)r -25 

H 

/ t 


\-6--4 -2/ 


2 i ! x 


\ 


; 

/ 
/ 



21. 



22. 



23. 



25. 





? y = 2x 




1 


2 

NX 1 


2a 


^^> 




-3 -2 -1 


V i ,/'* ' 


-1 


u,-i\ 


-2 


\ 20 

y + l-(x-l) 



24. 




26. 




Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 



30 

27. 



29. 



Chapter 1 Preliminaries 





28. 





j 

:4 
:3 

I / 

l/l 


IT 

; 1 

V. 


-6 -y - 


2 V 


2 4 6 


.. . M 


-2 





y + 1 = 



(x + 2) 



30. 




31. 



32. 




2 4 6 




33. 



34. 



35. 











3 








2 
1 


- *(i. i) 


1+1 


/jc-1 





1 2 




5 





36. 




Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle 



37. 




Section 1.5 Combining Functions; Shifting and Scaling Graphs 31 

38. 




39. 



40. 





41. 



42. 



1 










2 


" 


; 


v = 


1 
x-1 


1 




j 






__0 
-1 




2 


3 


4 


-2 


- 


\ 







7.5 

5 

2.5 


l-H 


-4 -2 

-2 


\^2 4 


^2.5 





43. 



y = - x +l 



12 3 



44. 



-7 



y = 



X + 2 



2 4 



Copffigl (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle 



32 
45. 



Chapter 1 Preliminaries 



1 . 


4 


-| ; l i 


3 


-/ ; 1 c»-i) 


2 
U 




-1 


1 2 3 



46. 



1 ) 

5 

/ 


1 

'-7-' 


_/j p — • 


^-~? & 


-1 

-5 

-10 





47. 





j 


« 






p 








/ 


_ 1 




_-' 


A 


- 








-2 - 


1 




1 2 



48. 



• -5 



y = (« + D 8 



2 4 



49. (a) domain: [0,2]; range: [2,3] 





> 






3 




\ y=/W + 2 




2 








1 













2 3 4 





(c) domain: [0,2]; range: [0,2] 



(b) domain: [0,2]; range: [-1,0] 



y = /W-l 




(d) domain: [0,2]; range: [—1,0] 




y = 2f(x) 



(e) domain: [—2,0]; range: [0,1] 





» 






t 


< 






2 






y=f(pc + 2) 

.f: 


1 






-2 -1 










y=-m 




(f) domain: [1,3]; range: [0,1] 





> 










2 












1 




l/ 




-1) 









1 


2 


3 





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Section 1.5 Combining Functions; Shifting and Scaling Graphs 
(g) domain: [—2,0]; range: [0,1] (h) domain: [— 1, 1]; range: [0,1] 



33 



y=f(-x) 




y=-f(x + l) + l 




50. (a) domain: [0,4]; range: [-3,0] 



(b) domain: [-4,0]; range: [0,3] 





«-t 



(c) domain: [—4,0]; range: [0,3] 




(d) domain: [-4,0]; range: [1,4] 



y = l -g(t) 



-2 




(e) domain: [2,4]; range: [—3,0] 



y g(-t + 2) 




(f) domain: [-2,2]; range: [-3,0] 




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34 Chapter 1 Preliminaries 



(g) domain: [1,5]; range: [—3,0] 




51. y = 3x 2 -3 

52. y = (2x) 2 - 1 = 4x 2 - 1 

53- y=K 1 +i)=l+2? 

54 -y = 1 + ^? = 1 + p 

55. y = V4X+1 

56. y = 3v/x+ 1 



57- y = \/4 -(f) 2 = |v / T6^ 



58. y= i\/4~ 



59. y = 1 - (3x) 3 = 1 - 27x 3 



60. y = i- (|) 3 = i- 



(h) domain: [0,4]; range: [0,3] 
y 





i 




3 




\ y--g(t-4) 





2 


4 



61. Lety = - V / 2x+l = f(x) and let g(x) = x 1 / 2 , h(x) = (x + 5) 1/2 ,i(x) = a/2(x + i) 1/2 ,and 

j(x) = — v 2(x +5) = i(x). The graph of h(x) is the graph of g(x) shifted left | unit; the graph of i(x) is the graph 
of h(x) stretched vertically by a factor of y 2; and the graph of j(x) = f(x) is the graph of i(x) reflected across the x-axis. 




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Section 1.5 Combining Functions; Shifting and Scaling Graphs 35 

62. Let y = ^/T^\ = f(x). Let g(x) = (~x) 1/2 , h(x) = (-x + 2) 1/2 , andi(x) = 4- (-x + 2) 1/2 = ^l^~\ = f(x). 

The graph of g(x) is the graph of y = ^fx reflected across the x-axis. The graph of h(x) is the graph of g(x) shifted right 
two units. And the graph of i(x) is the graph of h(x) compressed vertically by a factor of y 2. 





-3 -2-1 12 



63. y = f(x) = x 3 . Shift f(x) one unit right followed by a shift two units up to get g(x) = (x — 1) +2. 




64. y = (1 - xf + 2 = -[(x - l) 3 + (-2)] = f(x). Let g(x) = x 3 , h(x) = (x - l) 3 , i(x) = (x - l) 3 + (-2), and 

j(x) = — [(x — 1) + (—2)]. The graph of h(x) is the graph of g(x) shifted right one unit; the graph of i(x) is the graph of 
h(x) shifted down two units; and the graph of f(x) is the graph of i(x) reflected across the x-axis. 



g(x) 



h(x) 







10 




/y = x 






5 






-3 


-2 


-5 
-10 


1 


2 3 







10 


y = (x-l) 3 / 






5 


J 


-3 


-2 


-1 / 

/ 5 


12 3 






'-10 







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36 Chapter 1 Preliminaries 

65. Compress the graph of f(x) = j horizontally by a factor of 2 to get g(x) = ^- . Then shift g(x) vertically down 1 unit to 
geth(x) = i.-l. 




66. Letf(x) = i and g(x) = £ + 1 = * + 1 



— ^—2 + 1 = j-. — 1 2 + 1- Since v/2 « 1.4, we see that the graph of 
f(x) stretched horizontally by a factor of 1.4 and shifted up 1 unit is the graph of g(x). 





67. Reflect the graph of y = f(x) = ^fk across the x-axis to get g(x) = — -y/x. 





> 












4 


- 










3 


- 










2 


- 








-4 -3 -2 


-1 
-1 




1 


2 3 4 






-2 
-3 




y 


= -fc 






-4 











68. y = f(x) = (-2x) 2/3 = [(-l)(2)x] 2 / 3 = (-l) 2/3 (2x) 2/3 = (2x) 2/3 . So the graph of f(x) is the graph of g(x) = x 2 / 3 
compressed horizontally by a factor of 2. 




f(x) 




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69. 




Section 1.5 Combining Functions; Shifting and Scaling Graphs 

70. 



37 




-3-2-1 12 3 



71. 9x 2 + 25y 2 = 225 => h + h = 1 



72. 16x 2 + 7y 2 = 112 =>• 



(Vr) 



2 I 42 







> 












v 










6 


- 










4 




9x 2 i 


25/ = 225 






2 








-6 


U4 


-2 
-2 

4 

-6 




2 


4 y 6 




73. 3x 2 + (y - 2) = 3 => 



(y-2) 2 



74. (x+l) 2 + 2y 2 = 4^ 



*-C-i). 



+ 



W 











4 


" 3x 2 + (y - 2) 2 = 3 




/ 3 






1 2 
\ 1 




-2 - 


1 


1 2 



(x + l) 2 +2y 3 = 4 




75. 3(x- l) 2 + 2(y + 2) 2 = 6 

(x-i) 2 [y-(-2)] 2 _-, 



76. 6(x+f) 2 + 9(y-i) 2 = 54 



[-(-§)] , (y-ff ! 



3 2 



W 




6(x + f) +9(y-^)"=54 



3(j:-ir + 2(y + 2) z = 6 




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38 Chapter 1 Preliminaries 

2 2 

77. fg + \ = 1 has its center at (0, 0). Shiftinig 4 units left and 3 units up gives the center at (h, k) = (—4, 3). So the 



equation is — — ^-^ — h " 32 ' 
from (-8, 3) to (0,3). 



(*- + ±Y , (y - 3)^ 
42 -1- 32 



1. Center, C, is (—4, 3), and major axis, AB, is the segment 



> 


. 


10 




(x + 4) 2 (y-3) 2 g 
16 9 




. L.J 




-10 -8 -6 -4 -2 


2 


-2 





78. The ellipse \ + £ = 1 has center (h, k) = (0, 0). Shifting the ellipse 3 units right and 2 units down produces an ellipse 



with center at (h, k) = (3, —2) and an equation tx 4 ; + ^ — ^ r ^ 



1. Center, C, is (3, —2), and AB, the segment from 



(3, 3) to (3, —7) is the major axis. 



3 


' ( 


x-3) : _(y + 2f 


3 
2 

1 


4^-^25 


-1 
-1 

-2 

-3 

-4 
-5 
-6 

-7 




2 3 4 )p 



79. (a) (fg)(-x) = f(-x)g(-x) = f(x)(-g(x)) = -(fg)(x), odd 
(b) ( f -) (-x) = ^ = -SL = _ ( i) (X ), dd 



(O (f)(^) = f^ = li } = -(f)W'°dd 



gx) 
-g(x) 

-g(x) _ 

f(x) " 



(d) f 2 (-x) = f(-x)f(-x) = f(x)f(x) = f 2 (x), even 

(e) g 2 (-x) = (g(-x)) 2 = (-g(x)) 2 = g 2 (x), even 

(f) (f o g)(-x) = f(g(-x)) = f(-g(x)) = f(g(x)) = (f o g)(x), even 

(g) (g o f)(-x) = g(f(-x)) = g(f(x)) = (g o f)(x), even 
(h) (f o f)(-x) = f(f(-x)) = f(f(x)) = (f o f)(x), even 

(i) (g o g)(-x) = g(g(-x)) = g(-g(x)) = -g(g(x)) = -(g o g)(x), odd 



80. Yes, f(x) = is both even and odd since f(— x) = = f(x) and f(— x) = 



-f(x). 



Copyright (c) 2006 Pearson Education 




Section 1.6 Trigonometric Functions 39 



81. (a) 



(b) 



(c) 




(d) 




82. 




(fog)(x) 



f(x) = x- 7 



1.6 TRIGONOMETRIC FUNCTIONS 



1. (a) s = r<9 = (10)(f ) =87rm 

2. 6 = s - = ±2s = fK radians and 5 -f f 1 ^) = 225° 

r 8 4 4 \ n / 



(b) S = r0 = (10)(110°) (^) 



HOtt _ 55ir 
18 9 



111 



3. 6> = 80° => 6> = 80° (^p) = f 



s = (6) (t^) = 8.4 in. (since the diameter = 12 in. =>- radius = 6 in.) 



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40 Chapter 1 Preliminaries 

4. d = 1 meter => r = 50 cm => 6> = f = | = 0.6 rad or 0.6 (±f£) « 34° 



6» 


— TT 


2tt 
3 





7T 

2 


3tt 
4 


sin 





x/3 

2 





1 


1 
7^ 


cos 


1 


1 
2 


1 





1 


tan 8 





V^ 





und. 


-l 


cot 8 


und. 


1 
75 


und. 





-l 


sec 6* 


1 


-2 


1 


und. 


-V~2 


esc 8 


und. 


2 


und. 


1 


\pi 



6. 



6» 


3tt 
2 


3 


7T 

6 


7T 

4 


5tt 
6 


sin 6 1 


1 


2 


1 

2 


1 
75 


1 

2 


cos 8 





1 
2 


75 

2 


1 
V~2 


75 

2 


tan# 


und. 


->/3 


1 
73 


1 


1 
73 


cot6> 





i 


-V3 


1 


-V* 


sec 8 


und. 


2 


2 

y5 


V^ 


2 

75 


esc 8 


1 


2 

75 


-2 


V^ 


2 



7. cos x = — j, tan x = — | 



sinx= 7i' cosx= 75 



9. sin x = — 7- , tan x = — y 8 



10. sin x = j| , tan x = — y 



11. sin x 



1 2 
-T- , COS X = t- 

75 75 



12. cos x 



75 



, tan x 



75 



13. 




period = 7r 



14. 




period = 4tt 



15. 



V 


' 
* 


3' = cos nx 


i 




1 ° 




\ ' / 


2 




-1 











period = 2 



16. 



1 

o 

-l - 



y ' cos y 



' 3tt n 

T 



period = 4 



... -\ 
y ■ cos x 



17. 




period = 6 



18. 



1 t 



i... 



Ot 



-■ 



1/ I 1 



period = 1 



y » - cos 2i« 

/TA A, 



"r2, ; \3 / \4/7"\5 



y • cos x 



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Section 1.6 Trigonometric Functions 41 



19. 




period = 2ir 



20. 




period = 2ir 



y sin (jt+|) 



21. 




period = 2ir 



22. 



\ 111 
\ hi 

\ ! ^ 



-i\ 



period = 27r 



23. period = |, symmetric about the origin 



24. period = 1 , symmetric about the origin 



w 


lilt.. 


111 


fll 




\ II 

\ \! 

\ ! v 



25. period = 4, symmetric about the y-axis 




26. period = 4tt, symmetric about the origin 

s 

j 
2^ 




-2n 





! 



2* 



-6|1 S * CSC 1 




27. (a) Cos x and sec x are positive in QI and QIV and 

negative in QII and QUI. Sec x is undefined when 
cos x is 0. The range of sec x is (— oo, — 1] U [1, oo); 
the range of cos x is [— 1, 1]. 



y = cos x 




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42 Chapter 1 Preliminaries 



(b) Sin x and esc x are positive in QI and QII and 

negative in QUI and QIV. Csc x is undefined when 
sinx is 0. The range of csc x is (— oo, — 1] U [1, oo); 
the range of sin x is [— 1, 1]. 



y = csc x y y = sin x 




28. Since cot x = — — , cot x is undefined when tan x = 

tan x 

and is zero when tan x is undefined. As tan x approaches 
zero through positive values, cot x approaches infinity. 
Also, cot x approaches negative infinity as tan x 
approaches zero through negative values. 



y » tan x 




29. D: -oo < x < oo; R: y = -1, 0, 1 



y = Lsin xj 

-2K ,* „-7T 1 



L. 



k' h - h ' IH-^ > ' i >> 




y =f sin x| 



31. cos (x — |) = cos x cos (— |) — sin x sin (— |) = (cos x)(0) — (sin x)(— 1) = sinx 

32. cos (x + |) = cos x cos (|) — sin x sin (|) = (cos x)(0) — (sin x)(l) = —sin x 

33. sin (x + |) = sin x cos (|) + cos x sin (|) = (sin x)(0) + (cos x)(l) = cos x 

34. sin (x — | ) = sin x cos ( — f) + cos x sin ( — § ) = (sin x )(0) + (cos x)(— 1) = —cos x 

35. cos (A - B) = cos (A + (-B)) = cos A cos (— B) - sin A sin (— B) = cos A cos B - sin A (-sin B) 

= cos A cos B + sin A sin B 

36. sin (A - B) = sin(A + (— B)) = sin A cos (-B) + cos A sin(-B) = sin A cos B + cos A(-sinB) 

= sin A cos B — cos A sin B 

37. If B = A, A - B = => cos (A - B) = cos 0=1. Also cos (A - B) = cos (A - A) = cos A cos A + sin A sin A 

= cos 2 A + sin 2 A. Therefore, cos 2 A + sin 2 A = 1 . 

38. If B = 2tt, then cos (A + 27r) = cos A cos 2n — sin A sin 2n = (cos A)(l) — (sin A)(0) = cos A and 

sin (A + 2ir) = sin A cos 2ir + cos A sin 2ir = (sin A)(l) + (cos A)(0) = sin A. The result agrees with the 
fact that the cosine and sine functions have period 2ir. 

39. cos (7T + x) = cos 7r cos x — sin 7r sin x = (— l)(cos x) — (0)(sin x) = —cos x 



Copffigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 



Section 1.6 Trigonometric Functions 

40. sin (2ir — x) = sin 2ir cos (— x) + cos (2tt) sin (— x) = (0)(cos (— x)) + (l)(sin (— x)) = —sin x 



41. sin(^ -x) 



42. cos(^+ X ; 



= sin (y) cos(— x) + cos (y) sin(— x) 



= (— l)(cos x) - 



= cos (y) cos x 



- sin (y ) sin x = (0)(cos x) — (— l)(sin x) 



(0)(sin 
= sin x 



43. sin ^1 = sin (| 

44. cos ^ =cos(| 



+ f ) = sin | cos | + cos | sin 

y — sin | sin y 



(— x)) = —cos : 



4 



y") = cos f cos I 



! = (f) (I) + (#)(#) = 



45. cos 



fL=cos(f-f)=cosfcos(-|)-sinfsin(-|) = (i)(f)-(f)(-f) = i±f 



46. sin 5| = sin (y' - |) = sin (^) cos (- |) + cos (^) sin I 



(*)(*) 






43 



47 Pnl; 2 tt _ l+cos(f) _ !+_# _ 2+y^ 
t/. cus g — 2 — 2 — 4 



4 o rnQ 2 jr. _ 1+cos (!f ) _ 1 + 4 _ 2+^3 
to. cus 12 — 2 — 2 — 4 



4 n „ irl 2 jr. _ l-c°s(ff) _ \-& _ 2-y/3 
*t7. sin 12 — 2 — 2 — 4 



50. sin 2 f = i=^M = h± = ^f 



51. tan(A + B) 



52. tan (A - B) = sin( /^ = ^acosb-cos Acosg 

v y cos (A— B) cos A cos B+sin A so B 



tan A— tan B 
1 +tan A tan B 



53. According to the figure in the text, we have the following: By the law of cosines, c 2 = a 2 +-b 2 2ab cos 8 

= 1 2 _|_ 1 2 _ 2 cos (A - B) = 2 - 2 cos (A - B). By distance formula, c 2 = (cos A - cos B) 2 + (sin A - sin B) 2 
= cos 2 A — 2 cos A cos B + cos 2 B + sin 2 A — 2 sin A sin B + sin 2 B = 2 — 2(cos A cos B + sin A sin B). Thus 
c 2 = 2 - 2 cos (A - B) = 2 - 2(cos A cos B + sin A sin B) => cos (A - B) = cos A cos B + sin A sin B. 

54. (a) cos(A — B) = cos A cos B + sin A sin B 

sin 9 = cos(f - 6>) and cos 9 = sin(f - 9) 

Let 9 = A + B 

sin(A + B) = cos If - (A + B)l = cos[(f - A) - b] = cos (§ - A) cos B + sin (§ - A) sin B 

= sin A cos B + cos A sin B 
(b) cos(A — B) = cos A cos B + sin A sin B 

cos(A - (— B)) = cos A cos (-B) + sin A sin (-B) 

=> cos(A + B) = cos A cos (— B) + sin A sin (— B) = cos A cos B + sin A (—sin B) 

= cos A cos B — sin A sin B 
Because the cosine function is even and the sine functions is odd. 

55. c 2 = a 2 + b 2 - 2ab cos C = 2 2 + 3 2 - 2(2)(3) cos (60°) = 4 + 9 - 12 cos (60°) =13-12 (|) = 7. 
Thus, c = s/l « 2.65. 



56. c 2 = a 2 + b 2 -2abcosC = 2 2 + 3 2 - 2(2)(3)cos(40°) = 13- 12 cos (40°). Thus, c = sjll- 12 cos 40° « 1.951. 



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44 Chapter 1 Preliminaries 



57. From the figures in the text, we see that sin B 



If C is an acute angle, then sin C 



g . On the other hand, 



if C is obtuse (as in the figure on the right), then sin C = sin (n — C) = \ . Thus, in either case, 
h = b sin C = c sin B => ah = ab sin C = ac sin B . 
By the law of cosines, cos C = 



u2 2 

2 ° b ~ c and cos B 



2_ b 2 



2ac 



interior angles of a triangle is ir, we have sin A = sin (tt — (B 



(?) 



a 2 +b 2 -c 2 



a 2 +c 2 -b 2 



v 2abc ; 



(2a 2 + b 2 - c 2 



Moreover, since the sum of the 
C)) = sin (B + C) = sin B cos C + cos B sin C 
2 -b 2 ) = P => ah = be sin A. 



2ab 2ac 

Combining our results we have ah = ab sin C, ah = ac sin B, and ah = be sin A. Dividing by abc gives 



_h 

be 



sin C 



sin B 
b 



law of sines 
58. By the law of sines 

Thus sin B = ^ ~ 0.982. 



sin A sin B 

2 — 3 



'3/2 



By Exercise 55 we know that c = yl. 



59. From the figure at the right and the law of cosines, 
b 2 = a 2 +^ 2 2(2a) cos B 



4-4a 



a 2 - 2a + 4. 



Applying the law of sines to the figure, 



sin B 
b 



■/2Z2 



y/3/2 
b 



2a + 4 = b 2 



3 2 
2 a 



| a. Thus, combining results, 

= 1 a 2 + 2a - 4 




=> = a 2 + 4a — 8. From the quadratic formula and the fact that a > 0, we have 

60. (a) The graphs of y = sin x and y = x nearly coincide when x is near the origin (when the calculator 
is in radians mode), 
(b) In degree mode, when x is near zero degrees the sine of x is much closer to zero than x itself. The 
curves look like intersecting straight lines near the origin when the calculator is in degree mode. 



61. A = 2,B = 2tt, C 



-7T,D 



1 





y 


= 2sin(x + n>- 






/ n \ 
/ 2 \ 

' -1 


% 

2 


/ 3ff \ 
/ 2 \ 


5ir 
2 




-3 











62. A= i,B = 2, C= 1,D 




-1 i y »-jSin (rtx - it) +i 



Copyright (c) 2006 Pearson Education 




Section 1.6 Trigonometric Functions 45 



63. A 



1,B = 4, C = 0,D 



y 

,, y = 


2 ■ ruts , l 
-5r sln lTj + 5? 




f\n 








1 1' 

% 








-1 


\ 1 


/ 3 \ 5 , 




X 









64. A= ^,B = L, C = 0,D 




^sm^.L>0 



65. (a) amplitude = |A| = 37 

(c) right horizontal shift = C = 101 



(b) period = |B| = 365 

(d) upward vertical shift = D = 25 



66. (a) It is highest when the value of the sine is 1 at f(101) = 37 sin (0) + 25 = 62° F. 
The lowest mean daily temp is 37(— 1) + 25 = — 12° F. 
(b) The average of the highest and lowest mean daily temperatures = — ^- — - = 25° F. 
The average of the sine function is its horizontal axis, y = 25. 



67-70. Example CAS commands: 
Maple 

f := x -> A*sin((2*Pi/B)*(x-C))+Dl; 

A:=3;C:=0;D1:=0; 

fjist := [seq( f(x), B=[l,3,2*Pi,5*Pi] )]; 

plot( f_list, x=-4*Pi..4*Pi, scaling=constrained, 

color=[red,blue,green,cyan], linestyle=[ 1,3,4,7], 
legend=["B=l","B=3","B=2*Pi","B=3*Pi"], 
title="#67 (Section 1.6)"); 
Mathematica 



Clear[a, b, c, d, f, x] 

f[x_]:=a Sin[27r/b (x - c)] + d 

Plot[f[x]/.{a -»• 3, b ->■ 1, c -»■ 0, d 



0}, {x, -47r,47r }] 



67. (a) The graph stretches horizontally. 




B 


= 2 


'Pi 


B 


= 1 




B 


= 3 


; Pi 



Copyright (c) 2006 Pearson Education 




46 Chapter 1 Preliminaries 

(b) The period remains the same: period 



| B |. The graph has a horizontal shift of \ period. 




68. (a) The graph is shifted right C units. 



C = 




(b) The graph is shifted left C units. 

(c) A shift of ± one period will produce no apparent shift. | C | = 6 

69. The graph shifts upwards | D | units for D > and down | D | units for D < 0. 



A 

TV > ' ,7V 

7 V v / ,7 1 


3 
6- 

4-_ 

\ 2-f 


> 

r\ 
• /~\ * 

'/ w 

7 V 

/ \* 

1 \ 


! \ ,D = 3 

N / 7 \» x / 7 


/ - 10 \J- 5 


V\ '/ 


-2 





70. (a) The graph stretches | A | units. 




(b) For A < 0, the graph is inverted. 



1.7 GRAPHING WITH CALCULATORS AND COMPUTERS 



1-4. The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the graphs and 
has little unused space. 



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Section 1 .7 Graphing with Calculators and Computers 47 



1. d. 



2. c. 



y ffx) = x'-4x 2 -4x + 16 



-5 -4 -3 -2 - 




12 3 4 5 

f(x) = x 4 -7x : +6x 




3. d. 



4. b. 



f(x) = 5 + 12x-x 3 





5-30. For any display there are many appropriate display widows. The graphs given as answers in Exercises 5—30 
are not unique in appearance. 



5. [-2, 5] by [-15, 40] 



6. [-4, 4] by [-4, 4] 





7. [-2, 6] by [-250, 50] 



8. [-1, 5] by [-5, 30] 



. 


) 


/Ct)=x 5 -5x 4 +10 




50 








-2 / 




iV 3 4 f 6 




/ -50 








/ -100 








/ -150 








/ -200 








J -250 









30 


. f(x) = 4x 3 -x 4 




20 






10 






V 1 


12 3' 


, 5 



Coppgl (c) 1 Pen Etation, k, publishing as Pearson Addison-Wesle 



48 Chapter 1 Preliminaries 

9. [-4, 4] by [-5, 5] 




11. [-2, 6] by [-5, 4] 




13. [-1,6] by [-1,4] 




15. [-3, 3] by [0,10] 




10. [-2, 2] by [-2, 







e 


f(x) = x : (6-x 3 ) 






e 


/"\ 






4 


/ \ 






\ 2 


:/ \ 


-2 


-1 




i I2 






-2 - 


L 1 



12. [-4, 4] by [-8, 







8 ' 








*\ 










2 






-4 y 


3 -2 


-i-2 


1 


2/3 4 






-4 
-6 
-8 : 




1/3 / 2 

y = x (x - 



14. [-1,6] by [-1,5] 




16. [-1,2] by [0,1] 




-5-4-3-2-1 12 3 4 5 



Coppgl (c) 1 Pen Etation, k, publishing as Pearson Addison-Wesle 



17. [-5,1] by [-5, 5] 



Section 1 .7 Graphing with Calculators and Computers 

18. [-5,1] by [-2, 4] 



49 



2 4 6 8 10 




x + 3 



19. [-4, 4] by [0, 3] 



20. [-5, 5] by [-2, 2] 





21. [-10, 10] by [-6, 6] 



22. [-5, 5] by [-2, 2] 



-10-8 -6 -4~\ 



/(*) = 



-fr 




23. [-6, 10] by [-6, 6] 



24. [-3, 5] by [-2, 10] 



1 


1 a ^ 6. 2 - 15.+ 6 
\ m= 4.2-10. 






"5 -2 

-4 
-6 


5 10 >X 




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50 Chapter 1 Preliminaries 

25. [-0.03, 0.03] by [-1.25, 1.25] 





v 








1.0 


/ \ y = sin 250.* 




0.5 


1 \ 


-0.02 1 




\ 0.02/ 



27. [-300, 300] by [-1.25, 1.25] 





1.0 


' v = cos (5o) 


-300 \ 




\ /300 >X 




1-0.5 






J -1.0 





29. [-0.25, 0.25] by [-0.3, 0.3] 




26. [-0.1, 0.1] by [-3, 3] 



y = 3cos60x 




28. [-50, 50] by [-0.1, 0.1] 



1 . f X 
y = — sin — 
10 ho 




30. [-0.15, 0.15] by [-0.02, 0.05] 



y = x" +;jcos lOOx 




31. x 2 + 2x = 4 + 4y - y 2 =>• y = 2 ± a/-x 2 - 2x - 
The lower half is produced by graphing 

y = 2 - \/-x 2 - 2x + 8. 



(x+lf + (y-2) z = 9 




32. y 2 — 16x 2 = 1 => y = ± \/l + 16x 2 . The upper branch 
is produced by graphing y = \J 1 + 16x 2 . 




Coppgl (c) 1 Pen %$A\ k, publishing as Pearson Addison-Wesle 



Section 1.7 Graphing with Calculators and Computers 51 



33. 



' /(x)=-tan2x 




34. 



10 f \ f(x) = 3cot|~ |+1 

6 
4 
2 



35. 



f f(x) = sin 2x + cos 3x 




36. 




-6 -4 \-2 





f ( x ) = sin 3 x 



2 \4 /6 




37. 



4 








3 






• 


2 

1 








»•*••••„ 


~~~. 


2 


3 4 5 6 


-1 




•. 




-2 




• 




-3 




• 


y = lT3 


-4 









38. 



* •! 



-0.4. -0.2 



!•-■ 



y = sm 



■ f 1 

sin — 

l x 



0.2 • 0.4 



39. 



40. 



1 
8 


. 


\ 1 


y=x [x\ 


6 


t 


5 


.• 


\ 4 


•■ 


\ 3 


- 


* 2 


/ 


-5 -4 -3 -2 -1 

-1 


12 3 4 5 


-2 





x'-l 




41. (a) y = 1059. 14x - 2074972 

(b) m = 1059.14 dollars/year, which is the yearly increase in compensation. 



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52 Chapter 1 Preliminaries 



(c) 




1985 1990 1995 2000 2005 



(d) Answers may vary slightly, y = (1059. 14) (2010) - 2074972 = $53, 899 

42. (a) Let C = cost and x = year. 

C = (7960. 71)x- 1.6 x 10 7 

(b) Slope represents increase in cost per year 

(c) C = (2637.14)x - 5.2 x 10 6 

(d) The median price is rising faster in the northeast (the slope is larger). 

43. (a) Let x represent the speed in miles per hour and d the stopping distance in feet. The quadratic regression function is 

d = 0.0866x 2 - 1.97x + 50.1. 
(b) 



. 


. 














500 
















400 
















300 
















200 
















100 




















20 


40 


60 


80 


100 





(c) From the graph in part (b), the stopping distance is about 370 feet when the vehicle is 72 mph and it is about 525 feet 
when the speed is 85 mph. 

Algebraically: d quadratic (72) = 0.0866(72) 2 - 1.97(72) + 50.1 = 367.6 ft. 
d qU adratic(85) = 0.0866(85) 2 - 1.97(85) + 50.1 = 522.8 ft. 

(d) The linear regression function is d = 6.89x - 140.4 => d linear (72) = 6.89(72) - 140.4 = 355.7 ft and 

diinear(85) = 6.89(85) — 140.4 = 445.2 ft. The linear regression line is shown on the graph in part (b). The quadratic 
regression curve clearly gives the better fit. 




44. (a) The power regression function is y = 4.44647x 0511414 . 



Copyright (c) 2006 Pearson Education 




Chapter 1 Practice Exercises 53 



(b) 




2 4 6 8 10 12 14 16 18 20 

(c) 15.2 km/h 

(d) The linear regression function is y = 0.913675x + 4.189976 and it is shown on the graph in part (b). The linear 
regession function gives a speed of 14.2 km/h when y = 11 m. The power regression curve in part (a) better fits the 
data. 

CHAPTER 1 PRACTICE EXERCISES 



1. 7 + 2x > 3 => 2x > -4 => x > -2 



_i i i i j i 



-6-4-2 2 4 



2. -3x< 10 => x > -f 



10 
3 



■4X 



3. i(x-l) < ±(x-2) => 4(x- 1) < 5(x-2) 



=>4x-4<5x-10=^6<x 



x-3 > 4+x 



3(x-3) > -2(4 + x) 



=4> 3x - 9 > -8 - 2x => 5x > 1 => x > | 



5. |x+l| = 7=> x+ 1 = 7 or -(x 4-1) = 7 => x = 6orx 



6. |y-3|<4^-4<y-3<4=^-l<y<7 



-i—i — I I I I A I I I — +x 
0123456789 



7. |l-f|>§=> l-l<-|orl 
=>- x > 5 or x < — 1 



x ^ 3 

2^2 



| < -§ or -| > i => -x < -5 or -x > 1 



8. ^±? <5=> -5 < ^±Z < 5=^ _i5 < 2x4-7< 15=^ -22 < 2x < 8=^ -11 < x< 4 

9. Since the particle moved to the y-axis, —2 4- Ax = => Ax = 2. Since Ay = 3Ax = 6, the new coordinates 

are (x 4- Ax, y 4- Ay) = (-2 4-2,5 4-6) = (0, 1 1). 

10. (a) 







10 - 


■ • 










• 


5 




• 


• 




-10 


-5 


-5 
-10 - 


• 


5 




10 



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54 Chapter 1 Preliminaries 



(b) line slope 
AB 
BC 
CD 
DA 
CE 

BD 



2-8 -6 

10-6 _ 4 _ 2 



2 -(-4) — 6 — 3 

6-(-3) _ _9_ _ 3 

-4-2 -6 2 

l-(-3) _ 4 _ 2 

8-2 6 3 



-4-f 

is vertical and has no slope 

(c) Yes; A, B, C and D form a parallelogram. 

(d) Yes. The line AB has equation y — 1 = — | (x — 8). Replacing x by y gives y = — 5 (t — 8) + 1 

= — | (— y) + l=5+l=6. Thus, E (y .6) lies on the line AB and the points A, B and E are collinear. 

(e) The line CD has equation y + 3 = — § (x — 2) or y = — |x. Thus the line passes through the origin. 



11. The triangle ABC is neither an isosceles triangle nor is it a right triangle. The lengths of AB, BC and AC are 

I / I q i 

V 53, v 72 and v 65, respectively. The slopes of AB, BC and AC are 5, — 1 and g, respectively. 



12. P(x, 3x + 1) is a point on the line y = 3x + 1. If the distance from P to (0, 0) equals the distance from P to 
(-3, 4), then x 2 + (3x + l) 2 = (x + 3) 2 + (3 - 3x) 2 => x 2 + 9x 2 + 6x + 1 = x 2 + 6x + 9 + 9 - 18x + 9x 2 



18x = 17 or x 



y = 3x + 1 = 3 (||) + 1 = ^. Thus the point is P 



'17 23 N 
. 18 ' 6 ) 



13. y = 3(x - 1) + (-6) => y = 3x - 9 

14. y = -i(x+l) + 2^y = -|x+ f 

15. x = 



16. m = -^_ 6 3] = =£■ = -2 => y = -2(x + 3) + 6 =» y = -2x 



17. y = 2 



18. m 



5 - 3 
-2 - 3 



J2_ 
-5 



-§^y=-f(x-3) + 3^y = -§x+f 



19. y = -3x + 3 

20. Since 2x — y = —2 is equivalent to y = 2x + 2, the slope of the given line (and hence the slope of the desired line) is 2. 

y = 2(x - 1) + 1 => y = 2x - 5 



2 1 . Since 4x + 3y = 1 2 is equivalent toy = — fx + 4, the slope of the given line (and hence the slope of the desired line) is 



•f(x-4)-12^y 



iv _ 20 
3 A 3 



22. Since 3x — 5y = 1 is equivalent to y = |x — 4, the slope of the given line is | and the slope of the perpendicular line is 
-I- y=-!(* + 2)-3^y=-fx-f 

23. Sinceix+ |y = 1 is equivalent to y = — |x + 3, the slope of the given line is — | and the slope of the perpendicular line 

is |. y = |(x + 1) + 2 => y = |x + | 



24. The line passes through (0, —5) and (3, 0). m : 



0-(-5) 5 

3-0 3 



=> y = | X - 5 



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Chapter 1 Practice Exercises 55 



25 . The area is A = 7r r and the circumference is C = 27r r. Thus, r=^=>A = 7r(;f- 



C\ 2 C 2 



2- 



v 2?r / ' 4?r ' 



26. The surface area is S = 4ttt 2 =>■ r = (jM . The volume is V = |tt r 3 =>■ r 



S \V2 



Substitution into the formula for 



surface area gives S = 47rr = 4tt [^ 



3V\ 2 / 3 



27. The coordinates of a point on the parabola are (x, x 2 ). The angle of inclination 9 joining this point to the origin satisfies 
the equation tan 9 = — = x. Thus the point has coordinates (x, x 2 ) = (tan 9, tan 2 ^). 

28. tan 9 = ^ = ^ => h = 500tan6> ft. 

run o u u 



29. 



-3 -2 -1 



y = x 



12 3 



30. 




Symmetric about the origin. 



Symmetric about the y-axis. 



31. 



32. 



y = x~ -2x-l 





Neither 

33. y(-x 

34. y(-x 

35. y(-x 

36. y(-x 

37. y(-x 

38. y(-x 

39. y(-x 



Symmetric about the y-axis. 



(-x) 2 + 1 = x 2 + 1 = y(x). Even. 

(-x) 5 - (-x) 3 - (-x) = -x 5 + x 3 + x = -y(x). Odd. 



1 — cos(— x) = 1 — cos x = y(x). Even. 



sec(— x) tan(— x) 



sin(-x) _ -sinx 
cos 2 (— x) cos 2 x 



-sec x tan x = — y(x). Odd. 



(-x) 4 +l _ x 4 +l 



x'+l 



(-x) 3 -2(-x) -x 3 +2x x 3 -2x 



-y(x). Odd. 



1 — sin(— x) = 1 -I- sinx. Neither even nor odd. 



-x + cos(— x) = — x + cos x. Neither even nor odd. 



40. y(-x) = V(- x ) 4 - ! = V 7 * 4 - 1 = y(x). Even. 



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56 Chapter 1 Preliminaries 

41 



42. 



43. 



44. 



45. 



46. 



47. 



50. 



51. 



52. 



The function is defined for all values of x, so the domain is (-co, oo). 
Since | x | attains all nonnegative values, the range is [—2, oo). 

Since the square root requires 1 — x > 0, the domain is [—oo, 1]. 
Since y 1 — x attains all nonnegative values, the range is [—2, oo). 

Since the square root requires 16 — x 2 > 0, the domain is [—4, 4]. 



For values of x in the domain, < 16 — x 2 < 16, so < y 16 — x 2 < 4. The range is [0, 4]. 

The function is defined for all values of x, so the domain is (— oo, oo). 
Since 3 2 ~ x attains all positive values, the range is (1, oo). 

The function is defined for all values of x, so the domain is (— oo, oo). 
Since 2e~ x attains all positive values, the range is (—3, oo). 

The function is equivalent to y = tan 2x, so we require 2x ^ y for odd integers k. The domain is given by x ^ ^f for 

odd integers k. 

Since the tangent function attains all values, the range is ( — oo, oo). 

The function is defined for all values of x, so the domain is (— oo, oo). 

The sine function attains values from —1 to 1, so —2 < 2sin(3x + 7r) < 2 and hence —3 < 2sin(3x + n) — 1 < 1. The 

range is [—3, 1]. 

The function is defined for all values of x, so the domain is (— oo, oo). 

The function is equivalent to y = y x 2 , which attains all nonnegative values. The range is [0, oo). 

The logarithm requires x — 3 > 0, so the domain is (3, oo). 
The logarithm attains all real values, so the range is (— oo, oo). 

The function is defined for all values of x, so the domain is (— oo, oo). 
The cube root attains all real values, so the range is (— oo, oo). 

The function is defined for —4 < x < 4, so the domain is [—4, 4]. 

The function is equivalent to y = y|x|, —4 < x < 4, which attains values from to 2 for x in the domain. The 

range is [0, 2]. 

The function is defined for — 2 < x < 2, so the domain is [—2, 2]. 
The range is [— 1, 1]. 



53. First piece: Line through (0, 1) and (1, 0). m = 
Second piece: Line through (1, 1) and (2, 0). m 

1 - x, < x < 1 

2 - x, 1 < x < 2 



o-i _ ^i _ 
1-0 " 1 " 
0-1 _ ^1 
2-1 1 



-l=^y=-x+l = l — x 
= -l=>y = -(x-l) + l 



-x+2 = 2 -x 



f(x) 



54. First piece: Line through (0, 0) and (2, 5). m = 
Second piece: Line through (2, 5) and (4, 0). m 



5-0 

2-0 ~~ 2 

- On! _ zj> 

~~ 4-2 ~~ 2 



' ^ y = |x 



(x-2)+5 



•|x+ 10= 10- f 



f(x) 



f x, < x < 2 

(Note: x = 2 can be included on either piece.) 



10-#, 2<x<4 



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Chapter 1 Practice Exercises 57 



55. (a) 

(b) 

(c) 
(d) 

56. (a) 

(b) 
(c) 

(d) 



:og)(-l) = f(g(-l)) = f( 7 =TTl) = f (!) = T 
,of)(2)=g(f(2))=g(i)= * 



2.5 V 5 



fof)(x)=f(f(x))=f(I) 



gog)(x) = g(g(x)) = g(;^=) 



i = x , x + 

l/x ' ' 



/x + 2 



\/7^2 +2 \/l + 2V^+2 



fog)(-l) = f(g(-l)) = f(-y=T+l) = f(0) =2-0 = 2 
gof)(2) = f(g(2)) = g(2 - 2) = g(0) = ^OTT = 1 
fof)(x) = f(f(x)) = f(2 - x) = 2 - (2 - x) = x 

°g)W = g(gW) = g(^Ti) = ^/^xTT + i 



57. (a) (fog)(x) = f(g(x)) = f(Vx + 2) = 2 - (V^+2) = -x, x > -2. 

of)(x) = f(g(x)) = g(2 - x 2 ) = v / (2-x 2 ) + 2 = v/4-x 2 

(b) Domain of fog: [—2, oo). 
Domain of gof: [—2, 2]. 

(c) Range of fog: (— oo, 2]. 
Range of gof: [0, 2]. 



59. 



58. (a) (fog)(x) = f(g(x)) = f(v / T^c) = TTr^t = </l 

(gof)(x)=f(g(x))=g(v^) = 

(b) Domain of fog: ( — oo, 1]. 
Domain of gof: [0, 1]. 

(c) Range of fog: [0, oo). 
Range of gof: [0, 1]. 




The graph of f2(x) = fi(|x|) is the same as the 
graph of fi(x) to the right of the y-axis. The 
graph of f2(x) to the left of the y-axis is the 
reflection of y = fi(x), x > across the y-axis. 



60. 




The graph of f2(x) = fi (|x|) is the same as the 
graph of fi(x) to the right of the y-axis. The 
graph of f2(x) to the left of the y-axis is the 
reflection of y = fi(x), x > across the y-axis. 



Copyright (c) 2006 Pearson Education 




58 
61. 



Chapter 1 Preliminaries 




i 




It does not change the graph. 



62. 




The graph of f2(x) = fi ( |x|) is the same as the 
graph of fi(x) to the right of the y-axis. The 
graph of f2(x) to the left of the y-axis is the 
reflection of y = fi(x), x > across the y-axis. 



63. 



64. 




The graph of f2(x) = fi(|x|) is the same as the 
graph of fi(x) to the right of the y-axis. The 
graph of f2(x) to the left of the y-axis is the 
reflection of y = fi(x), x > across the y-axis. 




The graph of f2(x) = fi (|x|) is the same as the 
graph of fi(x) to the right of the y-axis. The 
graph of f2(x) to the left of the y-axis is the 
reflection of y = fi(x), x > across the y-axis. 



65. 



66. 









\ y- 1* 3 | 


1 


I 




7 


y 






i 


.* 




i 




1 




.* 


'1 




3 






. y = x 







Whenever gi(x) is positive, the graph of y = g2(x) 
= |gi(x)| is the same as the graph of y = gi(x). 
When gi(x) is negative, the graph of y = g2(x) is 
the reflection of the graph of y = gi (x) across the 
x-axis. 




It does not change the graph. 



67. 




Whenever gi(x) is positive, the graph of y = g2(x) = |gi(x)| is 
the same as the graph of y = gi(x). When gi(x) is negative, the 
graph of y = g2(x) is the reflection of the graph of y = gi(x) 
across the x-axis. 



Copyright (c) 2006 Pearson Education 




Chapter 1 Practice Exercises 59 



68. 




Whenever gi(x) is positive, the graph of y = g2(x) = |gi(x)| is 
the same as the graph of y = gi(x). When gi(x) is negative, the 
graph of y = g2(x) is the reflection of the graph of y = gi(x) 
across the x-axis. 



69. 







y = cos 2x 






/ ° 


\ K I 
\ 2 / 


K \ 3ll I 

\ 2 / 


2x 




' 1 











period = ir 



70. 



-2u 




period = 4tt 



ysin-j 



71. 




period = 2 



72. 




period = 4 



73. 



y 








i 


v 








2 
1, 






;p = 2cos(x-|) 




Ik 

-1 




3 


5k\ An 
6 \ 3 


l\\K 
1 6 


-2 











period = 2ir 



74. 




period = 2-k 



y = 1 + sin tx+j) 



75. (a) sin B = sin f = \ = \ => b = 2 sin f = 2 (^) = y^- By the theorem of Pythagoras, 



a 2 + b 2 = c 2 



a= Vc 2 - b 2 = ^4-3 = 1. 



(b) sin B = sin ? 



_ b _ 2 
3 c c 



(?y-vs 



^ . Thus, a = Vc 2 - b 2 = ^(^) 2 -(2) 2 



4 _ 2 

5 " v^' 



76. (a) sin A = - =4> a = c sin A 

77. (a) tan B = ^ => a = -^ 

^ ' a tan K 



(b) tan A 



(b) sin A 



a = b tan A 



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60 



Chapter 1 Preliminaries 



78. (a) sin A 



(c) sin A 



Vc2-b2 



79. Let h = height of vertical pole, and let b and c denote the 
distances of points B and C from the base of the pole, 
measured along the flatground, respectively. Then, 
tan 50° = ^, tan 35° = jj, and b - c = 10. 

C ' D ' 

Thus, h = c tan 50° and h = b tan 35° = 

=> c tan 50° = (c + 10) tan 35° 

=> c (tan 50° - tan 35°) = 10 tan 35° 

=> c = , ' ' ari l 5 ° => h = c tan 50° 



(c+ 10) tan 35° 



tan 50°-tan 35° 
10 tan 35° tan 50" ^ 1 1 no rn 
tan 50 —tan 35 




fr — io -*r 



80. Let h = height of balloon above ground. From the figure at 
the right, tan 40° = \, tan 70° = \, and a + b = 2. Thus, 
h = b tan 70° => h = (2 - a) tan 70° and h = a tan 40° 
=> (2 - a) tan 70° = a tan 40° => a(tan 40° + tan 70°) 
= 2 tan 70° => a = , l!™ 70 ' => h = a tan 40° 



balloon 



tan 40°+tan 70° 
2 tan 70° tan 40° ^ i -j i, 

*. Ana i .. ana '" S - J i-.J Jvlll. 

tan 40 + tan 70 




81. (a) 



y 

2t 



\y ,i 



y *sin x + cos: 



/ 



(b) The period appears to be \-k. 

(c) f(x + 4tt) = sin (x + 4tt) + cos (^) = sin (x + 2tt) + cos (| + 2tt) 
since the period of sine and cosine is 27r. Thus, f(x) has period 47r. 



sin x 



cos 



82. (a) 




y-s1n 7 



(b) D = (-oo,0)U(0,oo);R= [-1,1] 

(c) f is not periodic. For suppose f has period p. Then f ( ^ + kp) =f(i) 
integers k. Choose k so large that ^- + kp > - 

f ( y + kp) = sin ( (1/2 x k J > which is a contradiction. Thus f has no period, as claimed. 



sin 2tt = for all 



< mk^v < n - Butthen 



Copyright (c) 2006 Pearson Education 




Chapter 1 Additional and Advanced Exercises 
CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES 



61 



1. (a) The given graph is reflected about the y-axis. 




(c) The given graph is shifted left 1 unit, stretched 
vertically by a factor of 2, reflected about the 
x-axis, and then shifted upward 1 unit. 



y = -2f(x+l)+l 
(-4, 1) 




(b) The given graph is reflected about the x-axis. 




(-3, 0) 



(d) The given graph is shifted right 2 units, stretched 
vertically by a factor of 3, and then shifted 
downward 2 units. 



? = 3/(x-2)-2 
4L- »(2,4) 



(-1,-3) 




(-1,-2) 



(3,-2) 



2. (a) 



(b) 



3 

« 2 » 

-3 7u/ \l/2 -3 

-2 ■■ 
-3 




3. There are (infinitely) many such function pairs. For example, f(x) = 3x and g(x) = 4x satisfy 
f(g(x)) = f(4x) = 3(4x) = 12x = 4(3x) = g(3x) = g(f(x)). 

4. Yes, there are many such function pairs. For example, if g(x) = (2x + 3) 3 and f(x) = x 1 / 3 , then 
(f o g)(x) = f(g(x)) = f ((2x + 3) 3 ) = ((2x + 3) 3 ) 1/3 = 2x + 3. 

5. If f is odd and defined at x, then f(— x) = — f(x). Thus g(— x) = f(— x) — 2 = — f(x) — 2 whereas 

- g (x) = -(f(x) - 2) = -f(x) + 2. Then g cannot be odd because g(-x) = -g(x) =>• -f(x) - 2 = -f(x) + 2 
=S> 4 = 0, which is a contradiction. Also, g(x) is not even unless f(x) = for all x. On the other hand, if f is 
even, then g(x) = f(x) — 2 is also even: g(— x) = f(— x) — 2 = f(x) — 2 = g(x). 

6. If g is odd and g(0) is defined, then g(0) = g(-0) = -g(0). Therefore, 2g(0) = =>• g(0) = 0. 



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62 Chapter 1 Preliminaries 



For (x, y) in the 1st quadrant, |x| + |y| = 1 + x 
<=> x + y=l+x O y=l. For (x, y) in the 2nd 

quadrant, |x| + |y| = x + 1 <^> — x + y = x + 1 
<=> y = 2x + 1. In the 3rd quadrant, |x| + |y| = x + 1 
<& -x — y = x 4- 1 <^> y = -2x - 1. In the 4th 

quadrant, [x| + |y| = x + 1 -o- x + (— y) = x + 1 
<=> y = — 1 . The graph is given at the right. 



|x| + \y\ =l+x 



8. We use reasoning similar to Exercise 7. 

(1) 1st quadrant: y+jy|=x+|x| 
<^> 2y = 2x <^> y = x. 

(2) 2nd quadrant: y + |y| = x + x 

<^> 2y = x + (-x) = O y = 0. 

(3) 3rd quadrant: y + |y| = x + |x| 

& y + (-y) = x + (-x) <^> o = o 

=>• all points in the 3rd quadrant 
satisfy the equation. 

(4) 4th quadrant: y + |y| = x + jx| 

<^> y + (— y) = 2x <^> = x. Combining 
these results we have the graph given at the 
right: 




9. By the law of sines, 

10. By the law of sines, 



>/3 

sin | 



sin A sin B 

a — b 



sin - 
~ b~ 



sin A sin B sin B 

a — b — 3 



y/3 sin (tt/4) _ v^ {^J 



sin (tt/3) 



VI 



lB = |sinf = |(f) 



3\/2 
8 



11. By the law of cosines, a 2 = b 2 + c 2 — 2bc cos A =>• cos A 



b 2 + c 2 -a 2 _ 2 2 + 3 2 - 2 2 _ 3 



2bc 



2(2)(3) 



12. By the law of cosines, c 2 = a 2 + b 2 - 2ab cos C = 2 2 + 3 2 - (2)(2)(3) cos £ = 4 + 9 - 12 (&\ 
= 13 - 6\fl =>■ c = v/l3 - 6-Jl, since c> 0. 



13. By the law of cosines, b 2 = a 2 + c 2 — 2ac cos B => cos B 



a 2 +c 2 -b 2 _ 2 2 +4 2 -3 2 
2ac (2)(2)(4) 



4+16-9 
16 



U. Since < B < tt, sin B = \/ \ - cos 2 B = J\- i§ = v^p = ^ 



16 



16 



14. By the law of cosines, c = a +b — 2ab cos C =4> cos C 



2 +b 2 -c 2 _ 2 2 +4 2 -5 2 _ 4+16-25 



2ab " (2)(2)(4) 



16 



- £. Since < C < tt, sin C = V T - cos 2 C = Jl - ^ = 4p. 



256 ~~ 16 



15. (a) sin 2 x + cos 2 x = 1 => sin 2 x = 1 — cos 2 x = (1 — cos x)(l + cos x) =4> (1 — cos x) 



l+COS X 



1— cos x sin x 



sin x 1+cosx 

(b) Using the definition of the tangent function and the double angle formulas, we have 



tan 2 (f) 



s,„ 2 a) 


2 


cos 2 (J) 


" 1+C „ S ( 2 (|)) 



1 —cos X 

1 +COS X 



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Chapter 1 Additional and Advanced Exercises 63 



16. The angles labeled 7 in the accompanying figure are 
equal since both angles subtend arc CD. Similarly, the 
two angles labeled a are equal since they both subtend 
arc AB. Thus, triangles AED and BEC are similar which 
implies ^ = 2ac a °^ b 
=>■ (a - c)(a + c) = b(2a cos 6 — b) 

ds e - b 2 

2ab cos 9. 




17. As in the proof of the law of sines of Section P. 5, Exercise 57, ah = be sin A = ab sin C = ac sin B 
=>■ the area of ABC = \ (base)(height) = i ah = i be sin A = i ab sin C = \ ac sin B. 



18. As in Section P.5, Exercise 57, (Area of ABC) 2 



\ (base) 2 (height) 2 - 4 



1 Q 2u2 _ 1 „2u2 



a^h 



a 2 b 2 sin J C 

a 2 + b 2 -c 2 



I a 2 b 2 (1 — cos 2 C) . By the law of cosines, c 2 = a 2 + b 2 — 2ab cos C => cos C — — — 



Thus, (area of ABC) 2 = \ a 2 b 2 (1 - cos 2 C) = \ a 2 b 2 I 1 



.I 2 + b 2 - c 2 
2ab 



)*)-*{ 



1 



(a 2 + b 2 -c 2 ) 2N 



4a 2 b 2 



i (4a 2 b 2 - (a 2 + b 2 - c 2 ) 2 ) = i [(2ab + (a 2 + b 2 - c 2 )) (2ab - (a 2 + b 2 - c 2 ))] 

^ [((a + b) 2 - c 2 ) (c 2 - (a - b) 2 )] = i [((a + b) + c)((a + b) - c)(c + (a - b))(c - (a - b))] 



[( 



a + b + c'l 



-a + b + c\ I a-b + c^ I a + b - c 



2 



2 



> 



)] = s(s — a)(s — b)(s — c), where s 



a+b+c 



2 



Therefore, the area of ABC equals ^/s(s — a)(s — b)(s — c) . 



19. 1. b + c — (a + c) = b — a, which is positive since a < b. Thus, a + c < b + c. 

2. b — c — (a — c) = b — a, which is positive since a < b. Thus, a — c < b — c. 

3. c > and a<b => c — = c and b — a are positive =>• (b — a)c = be — ac is positive =>• ac < be. 

4. a < b and c < =>• b — a and — c are positive =^ (b — a)(— c) = ac — be is positive => be < ac. 

5. Since a > 0, a and - are positive =4> - > 0. 

6. Since < a < b, both i and 5 are positive. By (3), a < b and \ > => a (i) < b (j) or 1 < ^ 

^ 1 (s)<a(5) b y(3)sincei>0 =* i<i. 

7. a < b < =>■ i and 5 are both negative, i.e., \ < and 1 < 0. By (4), a < b and \ < => b Q) < a (^) 



^ ;<l =► i(5)<Ms) b y(4)^ei<0 



1 < i 
b V a 



20. (a) If a = 0, then = |a| < |b| <^ b ^ o 



|a| 2 < Ibl 2 . Since lal 2 



a 2 and 



b 2 we obtain a 2 < b 2 . If a ^ then |a| > and |a| < |b| ^> a 2 < b 2 . On the other hand, 



if a 2 < b 2 then a 2 



|a| z < Ibl" 



0<|b| 



(|b|-|a|)(|b| + |a|). Since (|b| + |a|) > 



and the product (|b| — |a|) (jbj + |a|) is positive, we must have (|b| — |a|) > =>• jbj > |a| . Thus 
|a| < |b| O a 2 < b 2 . 
(b) ab < |ab| => -ab > -2 |ab| by Exercise 19(4) above =^ a 2 - 2ab + b 2 > |a| 2 - 2 |a| |b| + |b| 2 , since 
|a| 2 = a 2 and |b| 2 = b 2 . Factoring both sides, (a - b) 2 > (|a| - |b|) 2 => |a-b| > ||a| - |b|| , by part (a). 



21. The fact that |ai + a2 + ... + a n | < |ai| + |a.2 1 + ••• + |a„| holds for n = 1 is obvious. It also holds for 
n = 2 by the triangle inequality. We now show it holds for all positive integers n, by induction. 

Suppose it holds for n = k > 1: |ai + &2 + ■ ■ ■ + a k | < |ai| + |a2 1 + • ■ • + Kl (this is the induction 
hypothesis). Then |ai + a 2 + ... + a k + a t+1 [ = |(ai + a 2 + ... + a k ) + a k+1 j < jai + a 2 + ... + a k | + |a k+[ | 
(by the triangle inequality) < |ai | + |a 2 1 + . . . + |a k | + |a k+1 1 (by the induction hypothesis) and the 
inequality holds for n = k + 1 . Hence it holds for all n by induction. 

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64 Chapter 1 Preliminaries 

22. The fact that |ai + a 2 + ... + a n | > |ai| — |&2 1 — ••• — |a n | holds for n = 1 is obvious. It holds for n = 2 
by Exercise 21(b), since |ai + a 2 | = |ai - (-a 2 )| > ||ai| - |-a 2 || = ||ai| - |a 2 || > |ai| - |a 2 | . 

We now show it holds for all positive integers n by induction. 

Suppose the inequality holds for n = k > 1. Then |ai + a 2 + • • • + a k | > |ai| — |a 2 | — . . . — |a k | (this is 
the induction hypothesis). Thus |ai + ... + a k + a k+1 1 = |(ai + . . . + a k ) — (— a k+1 )| 

> ||(ai + ... +a k )|-|-a k+1 || (by Exercise 21(b)) = ||ai + ... +a k |-|a k+l || > |ai+... + a k |-|a k+l 

> |ai| — |a 2 | — ... — |a k | — |a k+1 | (by the induction hypothesis). Hence the inequality holds for all 
n by induction. 

23. If f is even and odd, then f(— x) = — f(x) and f(— x) = f(x) =>• f(x) = — f(x) for all x in the domain of f. 
Thus 2f(x) = => f(x) = 0. 

24. (a) As suggested, letE(x) = f(x) + f( ~ x) => E(-x) = «-») + 3-(-»» = f(x) + f( ~ x) = E(x) => E is an 

even function. Define O(x) = f(x) - E(x) = f(x) - f(x)+ 2 f( ~ x) = f(x) ~ f( ~ x) . Then 
O(-x) = f( - x) - 2 ( - ( - x)) = ^^ = - { M ^ 1 ) = -0(x) ^ O is an odd function 

=> f(x) = E(x) + O(x) is the sum of an even and an odd function, 
(b) Part (a) shows that f(x) = E(x) + O(x) is the sum of an even and an odd function. If also 

f(x) = Ei(x) + Oi(x), where Ej is even and Oi is odd, then f(x) — f(x) = = (Ej(x) + Oi(x)) 
— (E(x) + O(x)). Thus, E(x) — E^x) = Oi(x) — O(x) for all x in the domain of f (which is the same as the 
domain of E — Ej and O — Oi). Now (E — Ei)(— x) = E(— x) — Ei(— x) = E(x) — Ei(x) (since E and Ej are 
even) = (E - Ej)(x) => E - Ei is even. Likewise, (Oi - 0)(-x) = Oi(-x) - O(-x) = -Oi(x) - (-O(x)) 
(since O and Oi are odd) = -(Oi(x) - 0(x)) = -(Oi - 0)(x) =4> Oi - O is odd. Therefore, E - Ej and 
Oi O are both even and odd so they must be zero at each x in the domain of f by Exercise 23. That is, 
Ei = E and Oi = O, so the decomposition of f found in part (a) is unique. 

25. y = ax 2 +bx + c = a(x 2 + ^x+|j) - g + c = a (x + ^) 2 - f d + c 

(a) If a > the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift 
of the vertex toward the y-axis and upward. If a < the graph is a parabola that opens downward. 
Decreasing a causes a vertical stretching and a shift of the vertex toward the y-axis and downward. 

(b) If a > the graph is a parabola that opens upward. If also b > 0, then increasing b causes a shift of the 
graph downward to the left; if b < 0, then decreasing b causes a shift of the graph downward and to the 
right. 

If a < the graph is a parabola that opens downward. If b > 0, increasing b shifts the graph upward 
to the right. If b < 0, decreasing b shifts the graph upward to the left. 

(c) Changing c (for fixed a and b) by Ac shifts the graph upward Ac units if Ac > 0, and downward —Ac 
units if Ac < 0. 

26. (a) If a > 0, the graph rises to the right of the vertical line x = — b and falls to the left. If a < 0, the graph 

falls to the right of the line x = — b and rises to the left. If a = 0, the graph reduces to the horizontal 
line y = c. As |a| increases, the slope at any given point x = Xo increases in magnitude and the graph 
becomes steeper. As |a| decreases, the slope at Xo decreases in magnitude and the graph rises or falls 
more gradually. 

(b) Increasing b shifts the graph to the left; decreasing b shifts it to the right. 

(c) Increasing c shifts the graph upward; decreasing c shifts it downward. 

27. If m > 0, the x-intercept of y = mx + 2 must be negative. If m < 0, then the x-intercept exceeds \ 

=>■ = mx + 2 and x > | =>■ x = - ^ > i =4> 0>m>-4. 



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Chapter 1 Additional and Advanced Exercises 65 



28. Each of the triangles pictured has the same base 
b = vAt = v(l sec). Moreover, the height of each 
triangle is the same value h. Thus h (base)(height) = \ bh 
= Ai = A2 = A3 = . . . . In conclusion, the object sweeps 
out equal areas in each one second interval. 




'a+0 b+(p 
v 2 ' 2 I 



'a M 
v2 ' 2) 



Thus the slope 



29. (a) By Exercise #95 of Section 1.2, the coordinates of P are 

Ax a/2 a 

(b) The slope of AB = §5^ = — - . The line segments AB and OP are perpendicular when the product 



of their slopes is -1 = (2) (- \) = 
is perpendicular to OP when a = b. 



- K ■ Thus, b 



2 



a = b (since both are positive). Therefore, AB 



Copyright (c) 2006 Pearson Education 




66 Chapter 1 Preliminaries 
NOTES: 



Copffigl (c| 1 Pearson Educate to, publishing as Pearson Addison-Wesle 



CHAPTER 2 LIMITS AND CONTINUITY 



2.1 RATES OF CHANGE AND LIMITS 

1 . (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x) 

approaches 1 . There is no single number L that all the values g(x) get arbitrarily close to as x — > 1 . 

(b) 1 

(c) 

2. (a) 

(b) -1 

(c) Does not exist. As t approaches from the left, f(t) approaches — 1. As t approaches from the right, f(t) 
approaches 1 . There is no single number L that f(t) gets arbitrarily close to as t — » 0. 



3. (a) True 
(d) False 



(b) True 

(e) False 



(c) False 
(f) True 



4. (a) False 
(d) True 



(b) False 
(e) True 



(c) True 



5. lim A does not exist because A = - = lifx>0 and A = — = — 1 if x < 0. Asx approaches from the left, 

x > Q |x| |x| x |x| —x rr 

A approaches — 1 . Asx approaches from the right, A approaches 1 . There is no single number L that all 
the function values get arbitrarily close to as x — > 0. 

6. As x approaches 1 from the left, the values of ^-r become increasingly large and negative. As x approaches 1 
from the right, the values become increasingly large and positive. There is no one number L that all the 
function values get arbitrarily close to as x — > 1, so lim -^-r does not exist. 

X — » 1 x ' 

7. Nothing can be said about f(x) because the existence of a limit as x — » xo does not depend on how the function 
is defined at Xo . In order for a limit to exist, f(x) must be arbitrarily close to a single real number L when 

x is close enough to Xo. That is, the existence of a limit depends on the values of f(x) for x near Xo, not on the 
definition of f(x) at x itself. 

8. Nothing can be said. In order for lim f(x) to exist, f(x) must close to a single value for x near regardless of 

x — > 

the value f(0) itself. 

9. No, the definition does not require that f be defined at x = 1 in order for a limiting value to exist there. If f(l) 
is defined, it can be any real number, so we can conclude nothing about f(l) from lim f(x) = 5. 

X — > 1 

10. No, because the existence of a limit depends on the values of f(x) when x is near 1, not on f(l) itself. If 



lim f(x) exists, its value may be some number other than f(l) 

X — > 1 



We can conclude nothing about lim f(x), 

X — » 1 



whether it exists or what its value is if it does exist, from knowing the value of f(l) alone. 




t(cl2 



na. dud Bima as 



Dc 




68 Chapter 2 Limits and Continuity 

11. (a) f(x) = (x 2 -9)/(x + 3) 



X 


-3.1 


-3.01 


-3.001 


-3.0001 


-3.00001 


-3.000001 


f(x) 


-6.1 


-6.01 


-6.001 


-6.0001 


-6.00001 


-6.000001 


X 


-2.9 


-2.99 


-2.999 


-2.9999 


-2.99999 


-2.999999 



f(x) 



5.9 



The estimate is lim f(x) 

x — > -3 



-5.99 
-6. 



-5.999 



-5.9999 -5.99999 -5.999999 



(b) 




f(x) = (x 1 - 9)/(x + 3) 



(c) f(x) = £=% = (x+ x 3 f 3 " 3) = x - 3 if x / -3, and _ lim „ (x - 3) = -3 - 3 = -6. 



x^ -3 



12. (a) g(x) = (x 2 - 2)1 (x - y/2) 



X 


1.4 


1.41 


1.414 


1.4142 


1.41421 


1.414213 


g(x) 


2.81421 


2.82421 


2.82821 


2.828413 


2.828423 


2.828426 



(b) 




(c) g(x) 



x^-2 



gOc) = (x 1 - 2)/(x - V2) 



U-V2 



a/2 if x ^ \/ 2 > and lim f x + \fl) = \Jl + \/2 = 2\/2. 

ti —* ,/9 V / 



^ 



13. (a) G(x) = (x + 6)/ (x 2 + 4x - 12) 



G(x) 



X 


-5.9 


-5.99 


-5.999 


-5.9999 


-5.99999 


-5.999999 


G(x) 


-.126582 


-.1251564 


-.1250156 


-.1250015 


-.1250001 


-.1250000 


X 


-6.1 


-6.01 


-6.001 


-6.0001 


-6.00001 


-6.000001 



-.123456 -.124843 -.124984 -.124998 -.124999 -.124999 



Copyright (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesley 



Section 2.1 Rates of Change and Limits 69 



(b) 



(c) G(x) 



10 



-10 



-20 



G(x) = (.x + 6)/(x 1 + ix-\2) 



x + 6 



x + 6 



(x 2 + 4x-12) (x + 6)(x-2) 



—^n if x ^ — 6, and lim — ^= 

x — 2 ' ' v , _c x — 2 



x^ -6 



-6-2 



-0.125. 



14. (a) h(x) = (x 2 -2x-3)/(x 2 -4x + 3) 



(b) 



h(x) 



X 


2.9 


2.99 


2.999 


2.9999 


2.99999 


2.999999 


h(x) 


2.052631 


2.005025 


2.000500 


2.000050 


2.000005 


2.0000005 


X 


3.1 


3.01 


3.001 


3.0001 


3.00001 


3.000001 



1.952380 1.995024 1.999500 1.999950 1.999995 1.999999 



y 

10 




i 


-l -» 

-10 
-20 




1 3 



(c) h(x) = i-j x 7\ = ( rl\^ + \\ = ^y if x ^ 3, and lim *±i = Ui 

v/ v/ x 2 — 4x + 3 (x — 3)(x— 1) x— 1 ' ' X ^Q x— 1 3 — 1 



15. (a) f(x) = (x 2 -l)/(|x|-l) 



X 


-1.1 


-1.01 


-1.001 


-1.0001 


-1.00001 


-1.000001 


f(x) 


2.1 


2.01 


2.001 


2.0001 


2.00001 


2.000001 



X 


-.9 


-.99 


-.999 


-.9999 


-.99999 


-.999999 


f(x) 


1.9 


1.99 


1.999 


1.9999 


1.99999 


1.999999 



(b) 




f(x) = (x 1 - l)/(|x| - 1) 



Copyright (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesley 



70 Chapter 2 Limits and Continuity 



(c) f(x) 



(x+lXx-1) 

x-1 

(x+lXx-1) 

-(x+1) 



x+1, x > Oandx + 1 

, and lim (1 - x) = 1 - (-1) = 2. 
1 — x, x < and x^-1 x — > - 1 



16. (a) F(x) = (x 2 + 3x + 2)/(2- |x|) 



X 


-2.1 


-2.01 


-2.001 


-2.0001 


-2.00001 


-2.000001 


F(x) 


1.1 


-1.01 


-1.001 


-1.0001 


-1.00001 


-1.000001 


X 


-1.9 


-1.99 


-1.999 


-1.9999 


-1.99999 


-1.999999 



F(x) 



-.99 



-.999 



-.9999 -.99999 -.999999 



(b) 



FM = t>r 2 + 3j: + 2V(2-W) 



(C) F(X) 




x >0 

rx + ^x + n > and lim (X+ 1) = -2+ 1 = -1. 

(x + 2)(x + i) _ x+1 x < Qandx ^ -2 * -» -2 



i7. (a) g(0) = (sin eye 



e 


.i 


.01 


.001 


.0001 


.00001 


.000001 


g(0) 


.998334 


.999983 


.999999 


.999999 


.999999 


.999999 


e 


-.1 


-.01 


-.001 


-.0001 


-.00001 


-.000001 



m 



.998334 .999983 .999999 .999999 .999999 .999999 



lim g(0) = 1 



(b) 



L — ,1^ 








3 


v- sin , ,. , 
\ y = — — (radians) 

\ 

\l U— \l L .1 


-5tt -Ait 


=3tt 


-2)h 


Sir 





TT^-^27T 37T 4-7T 57T 



NOT TO SCALE 



18. (a) G(t) = (l -cost)/t 2 



t 


.1 


.01 


.001 


.0001 


.00001 


.000001 


G(t) 


.499583 


.499995 


.499999 


.5 


.5 


.5 



t 


-.1 


-.01 


-.001 


-.0001 


-.00001 


-.000001 


G(t) 


.499583 


.499995 


.499999 


.5 


.5 


.5 



lim G(t) = 0.5 

t-t0 



Copyright (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesley 



Section 2.1 Rates of Change and Limits 7 1 



(b) 



H 1 H 



G(t) 



1 - cos t 



0.4 
4-0.3 

0.2 
0.1 



-H 1 1 

-0.0OO3 -0.0001 0.0001 0.0003 

Graph is NOT TO SCALE 

19. (a) f(x) = x 1 /' 1 ^) 



X 


.9 


.99 


.999 


.9999 


.99999 


.999999 


f(x) 


.348678 


.366032 


.367695 


.367861 


.367877 


.367879 


X 


1.1 


1.01 


1.001 


1.0001 


1.00001 


1.000001 



f(x) 



.385543 .369711 .368063 .367897 .367881 .367878 



lim f(x) w 0.36788 

X — > 1 



(b) 




C.9999 0.99995 



2.71815 

Graph is NOT TO SCALE. Also the intersection of the axes is not the origin: the axes intersect at the point 
(1,2.71820). 



20. (a) f(x) = (3 X - l)/x 



X 


.1 


.01 


.001 


.0001 


.00001 


.000001 


f(x) 


1.161231 


1.104669 


1.099215 


1.098672 


1.098618 


1.098612 


X 


-.1 


-.01 


-.001 


-.0001 


-.00001 


-.000001 



f(x) 



1.040415 1.092599 1.098009 1.098551 1.098606 1.098611 



lim f(x) w 1.0986 

x — > 



(b) 




21. lim 2x = 2(2) = 4 

x^2 



22. lim 2x = 2(0) = 

x — » 



23. lim i (3x-l) = 3(|) -1=0 



24. lim ^-r = -,,,, 1 , 

x _> ] 3x-l 3(1)- 1 



Copyright (c| 1 Pearson Education, k, publishing as Pearson Addison-Wesley 



72 Chapter 2 Limits and Continuity 



25. lim 3x(2x - 1) = 3(-l)(2(-l) - 1) 

X — > — 1 



26. lim &= ##, = -i 

x _» _J 2x— 1 2(— 1)— 1 -3 



27. lim x sin x = I sin I = 5 

Y , 7T Z Z 2 



28. lim f^ = <pJi = _=L = i 

X — > 7T 1— 7T 1 — 7T l—TT 7T — 1 



29. (a) M = f(^|2) = 28_9 = 19 



CM Af _ f(l)-f(-l) _ 2^0 _ -, 
{"' Ax ~~ l-(-l) ~ 2 ~ l 



in |" a ) A S - gci)-g(-i) _ 1-1 _ n 



CM ^£ — g( Q )-g(- 2 > _ 0^4 _ _ 9 
w Ax ~~ 0-(-2) ~ 2 ~ z - 



31. (a) 



Ah _ h(f)-Mf) _ -1-1 



At 



fU) Ah _ h(f)-h(f) _ Q-yf _ -3^3 
1 ' At - § - f - f - ,r 



32 (a) ^ = g (7r >~gW = (2-D-(2+D = _ 2 

' ^ ^ At 7T — 7T — 7T 



CM Ag _ g(7r)-g(-7r) _ (2-l)-(2-l) _ n 

W At ~~ TT-(-TT) — 271 _ U 



oo AR _ R(2) - R(Q) 
JJ - AS 2-0 



8+1- \/T 



3-1 
2 



o 4 AP _ P(2)-P(l) 
J ^' A0 2-1 



(8 — 16 + 10)— (1 -4 + 5) 

1 



2-2 = 



35. (a) 



Slope of PQ 



At 



Qi(10,225) 

Q 2 (14,375) 

Q 3 (16.5,475) 

Q 4 (18,550) 



650 - 225 
20- 
650- 



10 

375 



20-14 

650-475 
16.5 
-550 



20- 
650 



42.5 m/sec 
45.83 m/sec 
50.00 m/sec 
50.00 m/sec 



(b) At t = 20, the Cobra was traveling approximately 50 m/sec or 180 km/h. 



36. (a) 



Slope of PQ 



Ap 
At 



Qi(5,20) 

Q 2 (7,39) 

Q 3 (8.5,58) 

Q 4 (9.5,72) 



SO- 


-20 


10 


-5 


so- 


-39 


ld 


-7 


80- 


-58 


10- 


-S.5 


80- 


-72 



10-9.5 



12 m/sec 
13.7 m/sec 
14.7 m/sec 
16 m/sec 



(b) Approximately 1 6 m/sec 



37. (a) 




91 


92 93 




Year 


174- 


-62 112 


1994- 


1992 2 



56 thousand dollars per year 



(c) The average rate of change from 1991 to 1992 is 



Ap 



62-27 



At 



The average rate of change from 1992 to 1993 is— 



1992-1991 
1 1 1 - 62 



35 thousand dollars per year. 
'At | iw; _ |,„i — 49 thousand dollars per year. 
So, the rate at which profits were changing in 1992 is approximatley J (35 + 49) = 42 thousand dollars per year. 



Copyright (c| 1 Pearson Education, Inc., publishing as Pearson Addison-Wesley 



Section 2.1 Rates of Change and Limits 73 



38. (a) F(x) 
x 



(x + 2)/(x - 2) 
1.2 



1.1 



1.01 



1.001 



1.0001 



1 



F(x) -4.0 -3.4 

4 -°- ( 3 ' 5.0; 
-4.04; 
-4.0004 



-3.04 



AF 

Ax 

AF 
Ax 

Ax 



1.2-1 — 
-3.04 -(-3) _ 

1.01-1 
-3.0004 -(-3) 

1.0001- 1 



(b) The rate of change of F(x) at x = 1 is —4. 



AF 
Ax 

AF 

Ax 



3.004 -3.0004 -3 

- - - 3 '^ ( - 3) - -4.4; 



1.1 - 1 

-3.004- (-3) 
1.001- 1 



-4.004; 



39. (a) g = S^iffi 



2-1 
g(l+h)- 



gd) 



N/2-1 
2- 1 

v/l+h-1 



0.414213 



Ax (l+h)-l 

(b) g(x) = v /x 



Ag 
Ax 



(1-5) -g(l) 
1 .5 - 1 



1.5-1 
0.5 



0.449489 



1+h 




1.1 


1.01 


1.001 


1.0001 


1.00001 


1.000001 


\/l+h 


1.04880 


1.004987 


1.0004998 


1.0000499 


1.000005 


1.0000005 


(Vl+h- 


- l)/h 


0.4880 


0.4987 


0.4998 


0.499 


0.5 


0.5 



(c) The rate of change of g(x) at x = 1 is 0.5. 

'l+h-1 1 



(d) The calculator gives lim 
h^0 



40. (a) i) 



f(3)-f(2) 
3-2 



ii) ^P 



I _ 1 

T 2 

T-2 



21 2T __ ^— 1 

T-2 2T(T - 2) 



2-T 
-2T(2-T) 



_ J_ X ^ 2 

2T' L r * 



(b) 



(c) The table indicates the rate of change is —0.25 at t = 2. 



T 






2.1 


2.01 


2.001 


2.0001 


2.00001 


2.000001 


f(T) 


0.476190 


0.497512 


0.499750 


0.4999750 


0.499997 


0.499999 


(f(T) - 


- f(2))/(T - 


-2) 


-0.2381 


-0.2488 


-0.2500 


-0.2500 


-0.2500 


-0.2500 



(d) lim 

T^ 2 



J-) 
-2TV 



41-46. Example CAS commands: 
Maple : 

f := x -> (x A 4 - 16)/(x - 2); 

xO := 2; 

plot( f(x), x = x0-l..x0+l, color = black, 
title = "Section 2.1, #4 1(a)" ); 

limit( f(x), x = xO ); 
In Exercise 43, note that the standard cube root, x A (l/3), is not defined for x<0 in many CASs. This can be 
overcome in Maple by entering the function as f := x -> (surd(x+l, 3) — l)/x. 
Mathematica : (assigned function and values for xO and h may vary) 

Clear[f, x] 

f[x_]:=(x 3 - x 2 - 5x - 3)/(x + l) 2 

x0=-l;h= 0.1; 

Plot[f[x],{x,x0-h,x0 + h}] 

Limit[f[x],x — > xO] 



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74 Chapter 2 Limits and Continuity 

2.2 CALCULATING LIMITS USING THE LIMIT LAWS 



1. lim (2x + 5) = 2(-7) + 5 = -14 + 5 = -9 

x — > — 7 



2. lim (10 -3x)= 10-3(12)= 10-36= -26 



3. lim (-x 2 + 5x - 2) = -(2) 2 + 5(2) -2= -4+ 10 -2 = 4 



4. lim (x 3 - 2x 2 + 4x + 8) = (-2) 3 - 2(-2) 2 + 4(-2) 



8 = -16 



5. lim 8(t - 5)(t - 7) = 8(6 - 5)(6 - 7) = -8 
t — ► 6 



6. lim 3s(2s - 1) = 3 (|) [2 (f) - l] = 2 (f - f 



7. lim 

x^2 



x + 3 _ 2 + 3 _ 5 
x + 6 ~~ 2 + 6 ~~ 8 



8 - ^5 x^7 = 5^7 = -2 = "2 



y^-5 



■ 2 (-5) 2 _ 

u 5 5-y 5 -(-5) ~ 10 _ 2 



9 . urn X- - -<=£- -25-5 



10 lim y + 2 - 2 + 2 _ 4 A - i 

y ^2 y 2 + 5y + 6 (2)2 + 5(2) + 6 4+10 + 6 20 5 



11. lim 3(2x- l) 2 = 3(2(-l)- l) 2 = 3(-3) 2 = 27 

X — > — 1 



12. lim (x + 3) 1984 = (-4 + 3) 1984 = (-1) 1984 = 1 

x — > — 4 



13. lim (5-y) 4 / 3 = [5-(-3)] 4 / 3 = (8) 4 / 3 = ((8) 1 / 3 ) 4 = 2 4 = 16 

y — > -3 v ' 



14. lim (2z - 8) 1 / 3 = (2(0) - 8) 1 / 3 = (-8) 1 / 3 = -2 
z — > 



15. lim 



h^o \/3h+T+i V 3 <°) + 1 + 1 \A+i 



3 _ 3 



i(S lim 5 = 5 = 5 _ 5 

h^O v / 5h + 4 + 2 v/5(0) + 4 + 2 ,/4 + 2 4 



17. 1^ ^-i = lim 

h -► h h^O h 



y/jh + 1-1 y/3h+l + l 



V^+l = Um Ph+l^l = Hm 3h_ = Um 

V3h+1 + 1 h^0 h(y3h+l + lj h-»0 h^h+l + lj h -> 



/3h+l + l 



_J _ 3 

/T + l 2 



18. lim 

h->0 



y/5h + 4-2 
h 



lim 

h-»0 



lim 



(5h + 4)-4 



\/5h + 4-2 y/5h + 4 + 2 

h V5h + 4 + 2 ~~ i/To h(v/5h + 4 + 2) 



lim 



5h 



lim 



h( v /5h + 4 + 2) h^O \/5h + 4 + 2 



/i + 2 4 



19. lim 

x->5 



x-5 

x 2 -25 



lim 



lim 



5 (x + 5)(x-5) x *^ 5 x + 5 5+5 10 



20. lim 2 *t 3 _^ 

x —> — 3 x +4x + 3 



lim * + 3 _ lim -J— = I — - I 

x" -3 (x + 3)(x+l) x " -3 x + 1 -3 + 1 2 



21. lim 

x^ -5 



x + 5 



lim 

x^ -5 



(x + 5)(x-2) 

x + 5 



lim (x - 2) = -5 - 2 = -7 
-►-5 



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Section 2.2 Calculating Limits Using the Limit Laws 75 



22. lim x2 - 7x + 10 

x^2 x - 2 



lim. <»-/)(» -2) = lim (x - 5) = 2 - 5 = -3 



x^2 



x-2 



x^ 2 



23. lim 

t-> 1 



tl+1^1 _ Hm (t + 2)(t-l) = lim t + 2 _ 1+2 _ 3 



t 2 -l 



t_>l (t~D(t+l) t^i t+1 1 + 1 2 

lim <™±1> 



lim 



t+2 _ -1+2 



24 lim ' + 3t + 2 .,,,, ,,,,, 

t-T-l t'-t-2 ~~ t"-l (t-2)(t+l) ~~ t " "±i t-2 ~ -1-2 



25. lim t^4 = lim =$£2 = lim ^ = ^ = -\ 

x _> _2 xJ + 2x 2 x^-2 x 2 (x + 2) x ^ -2 x 4 2 



?fi lim ^ ^^ .MM MM, 

Z °- y™0 3y'-16y2 ~ /To y 2 (3y 2 -16) ~ ^"\ 3y 2 - 16 " -16 



lim 



lim 



5y + 8 _ _8_ 



j im (u 2 + l)(u+l)(u-l) _ lim (u 2 + l)(u+l) _ (1 + 1X1 + 1) _ 4 



27 Hm u l 

u"l u^-l-^-"! (u2 + u+l)(u-l) "u 1 ""! u2 + u + l 



lim 



(v - 2) (v 2 + 2v + 4) 



lim 



v 2 + 2v + 4 



4 + 4 + 4 _ 12 _ 3 



90 i; m v - 8 

Z0 - v" l 2 v 4 - 16 _ y 1 :^ (v-2)(v + 2)(v2 + 4) ~ v ^" 2 ( v + 2)(v2 + 4) " (4)(8) "32 



29. lim 

x^9 



xA-3 



lim 



xA-3 



lim 



l 



x " 9 x^9 (v /x - 3 )(v^ + 3 ) x^9 v^ + 3 v^ + 3 6 

x(2+y / x')(2-yx") 



30. lim £=£ = lim ^x) = lim 

x ^ 4 2 - V x x^4 2 - x /x x ^ 4 



2-, /x 



lim x (2 + a/x) = 4(2 + 2) = 16 



31. lim 



x-l 



lim 



_^_^_^^ - Hm i^M±3 + 2) = Um (V^T3 + 2 ) 



X ->1 v^T+3-2 X _>1 (V^+3-2) (VxT3 + 2) x _> 1 (x + 3)-4 x _> j 

= x/4 + 2 = 4 



32. lim 



V^T$-3 _ ,. (v / x^Ti-3)(7x2T8 + 3) 



x+1 



lim 



lim 



(x 2 + 8)-9 



1 (x + l)(Vx 2 + 8 + 3) x->-l (x+l)(V" 2 + 8 + 3) 



lim 



(x+l)(x-l) 



lim 



x-»-l (x+l)(Vx 2 + 8 + 3) x->-l \/x 2 + 8+3 3 + 3 



33. lim ^- 4 = lim. (^-*)J_g^ + *) = lim. 



(x 2 +12)-16 



x^2 



lim 

x^2 



x^2 



(x-2)(x + 2) 



(x-2) (Vx2 + 12 + 4) x^2 (x-2)(Vx 2 + 12 + 4) 



lim 



x + 2 



(x-2) (\7x 2 + 12 + 4) x^2 Vx 2 +12 + 4 \/l6 + 4 2 



34. lim 



x + 2 



2 V* 2 + 5 - 3 



lim 



lim 



(x + 2)(v'x 2 + 5 + 3) 
(7x2 + 5-3) (Vx 2 + 5 + 3) 

(x + 2)(y^TI + 3) _ ^5 + 3 _ 



(x + 2)(x-2) 



lim 



x-2 



= lim 

x^ -2 


(x + 2)(Vx 2 + 5 + 3j 


(x 2 + 5) - 9 


v/9 + 3 
-4 


3 

2 



35. lim 2 -vgp 

x^-3 x + 3 



lim 



(2 - v/ X 2 - 5) (2 + \/x 2 - 5) 

3 (x + 3)(2 + v/x 2 -5) x ^ -3 (x + 3)(2+ v / x 2 -5) 



lim 



4-(x 2 -5) 



lim 



9-x 2 



3 (x + 3)(2 + Vx 2 -5) 



lim 



(3-x)(3 + x) 



lim 



3 (x + 3)f2+^x 2 -5) x -> -3 2 + \/x 2 - 5 2 + v/4 



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76 Chapter 2 Limits and Continuity 



(4 - x) (5 + Vx2 + 9) . (4 - x) (5 + 7x2 + 9) 



36. lim — 4 , * n = lim -, — . ' v N , "- . ' s = lim ^ — 

x -> 4 5 - Vx2 + 9 x -> 4 (5 - 7x2 + 9 ) (5 + \/x^+9 ) x -> 4 25 - (x 2 - 



lim 

x^4 



(4 - x) (5 + Vx 2 + 9) (4 - x) (5 + \/x2 + 9) 

= lim ,. v ,,, . , — - = lim 

x -» 4 ( 4 ~ x )< 4 + x) x -> 4 



16 -x 2 



5 + ^x2 + 9 _ 5+y/25 _ 5 
4 + x ~~ 8 ~ 4 



37. (a 
(b 

(c 

38. (a 
(b 

(c 

39. (a 

(b 
(c 
(d 

40. (a 

(b 

(c 
(d 

41. (a 

(b 

(c 
(d 

42. (a 
(b 
(c 



quotient rule 

difference and power rules 

sum and constant multiple rules 

quotient rule 

power and product rules 

difference and constant multiple rules 



x lim c f( X )g(x) = [ x lim c f(x)j [ x lim g(x)] = (5)(-2) = -10 

x lim c 2f(x) g(x) = 2 [ x lim c f(x)] [ x lim c g(x)] = 2(5)(-2) = -20 
lim [f(x) + 3g(x)] = lim f(x) + 3 lim g(x) = 5 + 3(-2) = -1 



f(x) 



X — » c 

lim f(x) 



x — > c ' 
5 



x lml c f(x)-g(x) lim f(x) - lim g(x) 5-(-2) 7 



lim [g(x) + 3] = lim g(x) + lim 3 = -3 + 3 = 

x ^ 4 x ^ 4 x ^ 4 

lim xf(x) = lim x • lim f(x) = (4)(0) = 

x — > 4 x — > 4 x — > 4 

lim [g(x)] 2 = f lim g(x)l 2 = [-3] 2 = 9 

x — > 4 Lx — » 4 J 

lim -iW- = ^4 S(X) = ^1 = 3 

v -I A f(x) - 1 lim f(x) - lim 1 0-1 J 



lim [f(x) + g(x)] = lim f(x) + lim g(x) = 7 + (-3) = 4 

x — > b x^b x^b 

lim f(x) • g(x) = [ lim f(x)l [ lim g(x)l = (7)(-3) = -21 

x^b Lx^b JLx^b J 

lim 4g(x) = f lim 4J [ lim g(x)l = (4)(-3) = -12 

x — > b Lx^b J Lx^b J 

lim f(x)/g(x) = lim f(x)/ lim g(x) 

x — > b x^b x — » b 



J_ 
-3 



lim [p(x) + r(x) + s(x)] = lim p(x) + lim r(x) + lim s(x) = 4 + + (-3) = 1 

x — > — 2 x — > -2 x — > -2 x — » — 2 

x lim 2 p(x) • r(x) • s(x) = [ lim ^ p(x)] [^ lim ^ r(x)] ^ lim ^ s(x)] = (4)(0)(-3) = 
lim [-4p(x) + 5r(x)]/s(x) = [-4 lim p(x) + 5 lim r(x)| / lim s(x) = [-4(4) + 5(0)]/- 3 = f 



43. lim 



(l+h) 2 -l 2 



h^O n h^0 

(-2 + h)2-(-2)' 



lim 1+2h + h2 -' = lim ^±^ = lim (2 + h) = 2 



h^0 



h->0 



44. lim 
h->0 



lim 4 - 4h + h2 - 4 = lim ^^ = lim (h - 4) = -4 
h^0 h h^O h h -f 



45. lim [3(2 + h)- 4] -P(2) -4] = Hm 



3h 



h^0 



h->0 



46. lim ( - 2 H ( - 2) 
h^O h 



""' " Hm - 2 - ( - 2 + 1 ) = lim - h 



h^O 2h !>""') --'"-^ "' ~ h n h ' 4 -- h > 



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47. lim 



/7+h- y/7 
h 



lim 

h->0 



Section 2.2 Calculating Limits Using the Limit Laws 

(v^Th-V^)(v^Th+V^) _ (7 + h)-7 



77 



h v / 7+h+V7 



lim 

h^O hIJl + h- 



■Vt) 



lim 



lim 



h->0 h(/7+h+/7) h->0 x/v+h+v/? 2s/l 



h->0 h h->0 h(v/3h+l+ 1J h-»0 h ^3h + 1 + 1 



lim 



3h 



lim 



3 _ 3 



h( v / 3h+l + l) h->0 \/3h +1 + 1 2 



49. lim ^5 - 2x 2 = J 5 - 2(0) 2 = \/5 and lim v^ - x 2 = a/5 - (0) 2 = Jl; by the sandwich theorem, 

x ^ x^O 

lim f(x) = y^ 

x — ► 

50. lim (2 — x 2 ) =2 — = 2 and lim 2 cos x = 2(1) = 2; by the sandwich theorem, lim g(x) = 2 

x^O x^O x^O 



51. (a) lim ( 1 — *H =1 — 2 = 1 and lim 1 = 1; by the sandwich theorem, lim = x - sin x = 1 

x -> V 6 / 6 x -> J ' x _> 2-2 cos x 



(b) For x ^ 0, y = (x sin x)/(2 — 2 cos x) 
lies between the other two graphs in the 
figure, and the graphs converge as x — » 0. 



y = (x sin x)/(2 - 2 cos x) 



h(x) = 1 




52. (a) lim I k ~ ta) = lim h — lim ^ = ^—0=4 and lim i = i ; by the sandwich theorem, 

x^0V 224 /x^0 2 x^0 242 2 x->0 2 2 •* 



lim i^f^ = 1 

x -> x2 2 



(b) For all x/0, the graph of f(x) = (1 — cos x)/x 2 
lies between the line y = \ and the parabola 



1 „2 



x^/24, and the graphs converge as x — ► 0. 




2 24 



53. lim f(x) exists at those points c where lim x = lim x. Thus, c = c => c (1 — c) = 



X — » C X — > C 

,2 



c = 0, 1 , or — 1 . Moreover, lim f(x) = lim x = and lim f(x) = lim f(x) = 1 . 

x^0 x^O x ^ — 1 x^l 



54. Nothing can be concluded about the values of f, g, and h at x = 2. Yes, f(2) could be 0. Since the 
conditions of the sandwich theorem are satisfied, lim f(x) =—5^0. 

x^2 ~ 



55. 1 = lim 

x^4 



ftv-i e lim f(x) — lim 5 lim f(x) — 5 

^= X Hmx-lmT2 = ^^T~ =* lim f(x) - 5 = 2(1) => lim f(x) = 2 + 5 = 7. 

X— *4 X )4 A ' '4: A f l ± 



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78 Chapter 2 Limits and Continuity 



fix) 



56. (a) 1 = lim -j 

x — > -2 x 



lim f(x) lim f(x) 

x^-2 _ x^-2 
lim x 2 4 

x^-2 



lim f(x) = 4. 

x^ -2 



(b) 1= lim 

x — > -2 x 



ffx) 



lim 

x^ -2 



fix) 



lim 1 

x^-2 x 



lim 

x^ -2 



f(x) / l 



lim 

x^ -2 



ffx) 



-2. 



57. (a) = 3 - = [lim M=*\ [lim (x - 2)] = ^ [ 



f -f^- ) (x - 2) | = lim„ [f(x) - 5] = Jim„ f(x) - 5 



x^2 



x^2 



lim ffx) = 5. 

x^ 2 



(b) = 4 • = [ lim f -^5f} [ lim (x - 2)1 => lim f(x) = 5 as in part (a). 

La ' £ J La. ? Z. J X. f £* 



58. (a) = 1 • = lim |£ lim x 



x -> x J Lx -> 



Inn ':)- ' lim x 2 1 = lim | Sg • x 2 



lim f(x). That is, lim f(x) = 0. 

x — > x — > 



(b) = 1 • = [ lim ^1 [ lim xl = lim [^ - xl = lim ® . That is, lim ^ = 0. 

Lx -> x J Lx -> J x -> L x J x -> x x -> x 



59. (a) lim x sin ! = 

v ' x -> x 



\ ly — x sin • 




1 



(b) -1 < sini < 1 forx^O: 
x > =>• —x < x sin - < x 

x < => —x > x sin ; > x 



lim x sin - = by the sandwich theorem; 

x^O x J 

lim x sin - = by the sandwich theorem. 

x^O x J 



60. (a) lim x 2 cos (4) 

x — > vx 







(b) -1 < cos (^) < 1 forx ^ => -x 2 < x 2 cos (^) < x 2 



theorem since lim x 2 = 0. 

x^O 



2.3 PRECISE DEFINITION OF A LIMIT 



i. -i- 



^-^ 



0.4 /i(x) = x 2 cos(l/i 3 ) 




-0.4 



lim x 2 cos (4) = by the sandwich 

x — » v x 



Step 1: \x-5\<6^>-6<x-5<6^>-6 + 5<\<5 + 5 
Step 2: .5 + 5 = 7 =^> 6 = 2,or-6 + 5=l =5> 6 = 4. 

The value of S which assures x — 5| < <5 =$■ l<x<7is the smaller value, 6 = 2. 



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Section 2.3 Precise Definition of a Limit 79 



-+- 



-+X 



1 2 7 

Step 1: \x-2\<6=>-6<x-2<S=>-6 + 2<x<S + 2 
Step 2: -6 + 2=1 => 6 = 1, or 6 + 2 = 1 => 6 = 5. 

The value of 6 which assures |x — 2| < 6 => l<x<7is the smaller value, 6=1. 



+- 



->-^ 



-7/2 -3 -1/2 

Step 1: |x - (-3)| <8 => -6 < x + 3 < 6 => -6-3<x<<5-3 
Step 2: -6-3 = -|^>6=i,or6-3 = -i=^<5=|. 

The value of 6 which assures |x — (— 3)| < <5 => — |<x<— | is the smaller value, 6 = |. 



H ) »-x 

2 2 



Step 1: |x - (- |) | < 6 => -6 < x + | < 6 => -6 - § < x < <5 
Step 2: -6 - | = - | => <5 = 2, or 6 - § = - \ => 6 = 1. 
The value of 6 which assures |x - 



1)1 <« 



| < x < — | is the smaller value, 6 = 1. 



5. -i, 



+^x 



4/9 1/2 4/7 

Step 1: |x-i|<(5=>-(5<x-5<6^>-i5+i<x<6+i 
Step 2: -6 + | = I => 6 = i, or 6 + ± = f => 6 = i 



The value of 6 which assures |x — || < 6 => | < x < | is the smaller value, 6 = X . 



-f 



+- 



-) •"« 



2.7591 3 3.2391 

Step 1: |x-3|<<5=^-6<x-3<<5=^-6 + 3<x<6 + 3 
Step 2: -6 + 3 = 2.7591 => <5 = 0.2409, or 6 + 3 = 3.2391 => <5 = 0.2391. 

The value of 6 which assures |x — 3| < 6 =4> 2.7591 < x < 3.2391 is the smaller value, 6 = 0.2391. 

7. Step 1: |x-5|<<5=^-6<x-5<<5^--6 + 5<x<<5 + 5 

Step 2: From the graph, -6 + 5 = 4.9 =4> 6 = 0.1, or 8 + 5 = 5.1 => 6 = 0.1; thus 8 = 0.1 in either case. 

8. Step 1: |x - (-3)| <8 => -8 < x + 3 < <5 => -6-3<x<<5-3 

Step 2: From the graph, -6-3 = -3.1 => 6 = 0.1, or 6 - 3 = -2.9 => 8 = 0.1; thus 6 = 0.1. 

9. Step 1: |x - 1| < 8 => -8 < x - 1 < 8 =>- -8+ 1 < x < 8 + 1 

Step 2: From the graph, -8+1 = ^ => 8 = ^, or 6 + 1 = f| =>• 6 = ^; thus 6 = ^. 

10. Step 1: \\~3\<8=>-8<\-3<8^-8 + 3<x<8 + 3 

Step 2: From the graph, -6 + 3 = 2.61 => 6 = 0.39, or 8 + 3 = 3.41 => 8 = 0.41; thus 8 = 0.39. 

11. Step 1: |x-2|<<5^-6<x-2<<5^-6 + 2<x<<5 + 2 

Step 2: From the graph, -6 + 2=a/3 =4> 6 = 2 - a/3 « 0.2679, or 6 + 2 = a/5 => 6 = a/5 - 2 « 0.2361; 
thus 6 = a/5 - 2. 



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80 Chapter 2 Limits and Continuity 



12. Stepl: |x-(-l)|<<5 => -6<x+l<6 => -6-l<x<6-l 



Step 2: From the graph, —6—1 



/5-2 



0.1180, or 6- 1 = - 



A 



2-V3 



0.1340; 



thus (5 



/5-2 



13. Stepl: |x-(-l)|<<5 => -6<x+l<6 => -<5 - 1 < x < <S - 1 

Step 2: From the graph, -(5 - 1 = - f => § = | « 0.77, or 6 - 1 - lh 



25 "^ 25 



^ - 0.36; thus 5=4 = 0.36. 



25 



14. Step 1: |x- \\ < S =4> -6 < x - \ < 6 =^ 
Step 2: From the graph, — 6 + \ = jjjy =>- 6 
thus 6 = 0.00248. 



-6+±<x<6+| 



1 

2.01 



0.00248, or 6 + \ = ^ 



6 = m~ I ~ 0.00251; 



15. Step 1: |(x + 1) - 5| < 0.01 => |x - 4| < 0.01 => -0.01 < x - 4 < 0.01 => 3.99 < x < 4.01 
Step 2: |x - 4| < 6 => -6 < x - 4 < 6 => -6+4<x<6+4 => 6 = 0.01. 

16. Step 1: |(2x - 2) - (-6)| < 0.02 => |2x + 4| < 0.02 =^> -0.02 < 2x + 4 < 0.02 => -4.02 < 2x < -3. 

=> -2.01 <x < -1.99 
Step 2: |x - (-2)| < 6 => -6 < x + 2 < 6 => -6-2<x<6-2 => 6 = 0.01. 



17. Step 1: k/x+ 1 - 1 < 0.1 =>• -0.1 < a/x + 1 - 1 < 0.1 => 0.9 < a/x + 1 < 1.1 => 0.81 <x+ 1 < 1.21 

=> -0.19 <x < 0.21 
Step 2: |x - 0| < 6 => -6 < x < 6. Then, -5 = -0.19 ^ 6 = 0.19 or 6 = 0.21; thus, 6 = 0.19. 



18. Step 1: | y/x - 1 1 < 0.1 => -0.1 < y/x - \ < 0.1 => 0.4 < ,/x < 0.6 => 0.16 < x < 0.36 
Step 2: |x-±|<6=^-6<x-i<6=4--6+±<:r<6+i. 

Then, -6 + \ = 0.16 =^ 6 = 0.09 or 6 + \ = 0.36 ^ 6 = 0.11; thus 6 = 0.09. 



19. Stepl: k/l9-x-3 < 1 =^ -1< a/19-x-3 < 1 => 2 < a/19-x <4 

^ -4 > x - 19 > -16 =^> 15>x>3or3<x<15 
Step 2: |x - 10| < 6 => -6 < x - 10 < 6 => -6 + 10 < x < 6 + 10. 

Then -6+10 = 3 => 6 = 7, or 6 + 10 = 15 =4> 6 = 5; thus 6 = 5. 



=> 4< 19-x< 16 



20. Step 1: |a/x-7-4| < 1 =$► -1 < i/x - 7 - 4 < 1 => 3 < y/x - 7 < 5 => 9<x-7<25 => 16<x<32 

Step 2: |x - 23 1 < 6 => -6 < x - 23 < 6 => -6 + 23 < x < 6 + 23. 

Then -6 + 23 = 16 =>■ 6 = 7, or 6 + 23 = 32 => 6 = 9; thus 6 = 7. 



21. Step 1: |i - 1| < 0.05 => -0.05 < 1 - \ < 0.05 => 0.2 < ± < 0.3 = 
Step 2: |x - 4| < 6 =>• -6 < x - 4 < 6 => -6 + 4 < x < 6 ' + 4. 

Then -6 + 4 = y or 6 = §, or 6 + 4 = 5 or 6 = 1; thus 6=\. 



i| >x> i° or i° <x< 5. 



22. Step 1: |x 2 - 3| < 0.1 => -0.1 < x 2 - 3 < 0.1 =4> 2.9 < x 2 < 3.1 => \J2.9 < x < a/3.1 
Step 2: |x - 1/3 < 6 => -8 <x- y/3 < 8 => -6+y/3<x<6 + a/3. 

Then -6 + a/3 = a/2^9 =^> 6 = a/3 - a/Z9 w 0.0291, or 6 + a/3 = a/3T => 5 = 
thus 6 = 0.0286. 



3.1- a/3 w 0.0286; 



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Section 2.3 Precise Definition of a Limit 81 



23. Step 1: |x^ - 4| < 0.5 => -0.5 < x 2 - 4 < 0.5 ^ 3.5 < x 2 < 4.5 => V3-5 < |x| < V4.5 => - V4.5 < x < -^3.5 
for x near —2. 
Step 2: |x - (-2)| <6 => -<5 < x + 2 < ,5 => -<5-2<x<<5-2. 

Then -6 - 2 = -y^ =^> 6 = ^5 - 2 « 0.1213, or 6 - 2 = 
thus<5 = a/45 -2« 0.12. 



'3.5 => <5 = 2- V3.5« 0.1292; 



10 „„ 10 



24. Stepl: |*-(-l)| < 0.1 => -0.1 < i + 1< 0.1 => - ±± < ± < - ^ =*► - ±2 > x > - f or - f < x 



in 
n • 



Step 2: |x-(-l)| <6 => -6 < x+ 1 < 6 =4- -S-Kx<6-1. 

Then -S - 1 = - f =>■ <5=i,or<5-l = -{f =4> § = i; thus <5 = i 



25. Stepl: |(x 2 -5) - 11| < 1 => |x 2 - 16| < 1 =>■ -1< x 2 - 16 < 1 =^> 15 < x 2 < 17 =4- v^ < x < \/vi. 
Step 2: |x-4|<<5=^-<5<x-4<<5=>-<5 + 4<x<<5 + 4. 

Then -S + 4= v^ => <5 = 4 - v^ « 0.1270, or <5 + 4 = v^ =^ <5 = v^ - 4 w 0.1231; 
thus<5 = v^-4^0.12. 



26. Stepl: |^-5| < 1 => -1 < ±f - 5 < 1 => 4< ±f <6 => ?>ifo>s => 30>x>20or20<x<30. 
Step 2: |x - 24 1 < 6 => -S < x - 24 < S => -6 + 24 < x < 6 + 24. 

Then -<5 + 24 = 20 => 6 = 4, or 5 + 24 = 30 => 6 = 6; thus ^6 = 4. 

27. Step 1: |mx - 2m| < 0.03 =4> -0.03 < mx - 2m < 0.03 => -0.03 + 2m < mx < 0.03 + 2m =► 

2 _OXB <x<2+ 0JB^ 
m m 

Step 2: \x-2\<6^-6<x-2<S^-S + 2<x<6 + 2. 

Then -<5 + 2 = 2 - ^ =>. ( 5 = PJ»^ + 2 = 2+ o : 03 ^ ^ = O03 ^ ith g = O03 

m m ' m m ' m 

28. Step 1: |mx - 3m| <c =>• -c<mx-3m<c => -c + 3m < mx < c + 3m => 3-^<x<3 + ^ 
Step 2: |x-3|<<5^-<5<x-3<<5^-<5 + 3<.r<<5 + 3. 

Then -6 + 3 = 3-^ =^ 5 = -^, or 6 + 3 = 3+^ => 6 = ±, In either case, 8 = ±. 

1 m m " ' ' m m ' m 



29. Step 1: |(mx + b) - (f + b) | < c => -c < mx - f < c =>• -c+f<mx<c+f =>• 
Step 2: |x-i|< ( 5=>-(5<x-i<5^>- l 5+i<x<6+i. 

6=-,or<5+i = i + - =>- 5=-. In either case, 6 



2 m ^ X ^ 2 + m- 



Then -<5 + \ = \ - *■ 

2 2 m 



30. Step 1: |(mx + b) - (m + b)| < 0.05 => -0.05 < mx - m < 0.05 =>• -0.05 + m < mx < 0.05 + m 



1-^5 <x< 1 + ^. 



Step 2: \x—l\<6=>-6<x-l<6=>—6+l<x<6 + l. 

Then -6 + 1 = 1 - ^ => <5=^, r<S + l = l + 2:2§ =S> 6 = ^. In either case, 6 = 2^5. 



31. lim (3 - 2x) = 3 - 2(3) 

x — ► 3 



-0.02 < 6 - 2x < 0.02 =>• -6.02 < -2x < -5.98 =^ 3.01 > x > 2.99 or 



Stepl: |(3 -2x) - (-3)| < 0.02 : 

2.99 < x < 3.01. 
Step 2: 0<|x-3|<(5 => -6 < x - 3 < 6 => -<5 + 3<x<<5 + 3. 

Then -6 + 3 = 2.99 =>■ <5 = 0.01, or S + 3 = 3.01 =^ <5 = 0.01; thus S = 0.01. 



32. lim (-3x-2) = (-3)(-l)-2= 1 

X — > — 1 



Step 1: |(-3x - 2) - 1| < 0.03 => -0.03 < -3x - 3 < 0.03 => 0.01 > x + 1 > -0.01 => -1.01 < x < -0.99. 



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82 Chapter 2 Limits and Continuity 

Step 2: |x-(-l)|<<5 => -6<x+l<6 => -6-l<x<6-l. 

Then -6 - 1 = -1.01 => 6 = 0.01, or 6 - 1 = -0.99 => 5 = 0.01; thus S = 0.01. 



lim (x + 2) = 2 + 2 = 4, x ^ 2 

x^ 2 



33. 


lim — 

x^2 x " 


-4 
-2 


lim 

x^2 


(x + 2)(x - 2) 
(x-2) 




Step 1: 


1 V x 


*)• 


- 4 1 < 0.05 



-0.05 < 



(x + 2)(x-2) 
(x-2) 



4 < 0.05 =4> 3.95 < x + 2 < 4.05, x ^ 2 



=> 1.95 <x< 2.05, x^ 2. 
Step 2: |x - 2| < 6 => -S < x - 2 < <5 => -6 + 2 < x < 6 + 2. 

Then -8 + 2= 1.95 => <5 = 0.05, or 6 + 2 = 2.05 ^ 6 = 0.05; thus S = 0.05. 



34. lim 

x^ -5 



-6x + 5 



lim 



(x + 5)(x+l) 



lim (x + 1) = -4, x / -5. 



-6x + 5 '\ i a\\ ^ r\ r\c _>. nn<^ (x + 5)(x + 1) 



Step 1: I ( x2 tf 5 +5 ) - (-4)1 < 0.05 => -0.05 < 



, y - -.,, .,,, .,._. (x ,, +4 < 0.05 =>• -4.05 < x+ 1 < -3.95, x/ -5 

=> -5.05 <x < -4.95, x^ -5. 
Step 2: |x - (-5)| <6 => -S <\ + 5 <6 =4> -6-5<x<6-5. 

Then -6 - 5 = -5.05 ^ 6 = 0.05, or <5 - 5 = -4.95 => <5 = 0.05; thus <5 = 0.05. 



35. lim Vl -5x = y/l - 5(-3) = v^ = 4 



x^ -3 

Step 1: I a/I -5x - 4| < 0.5 => -0.5 < \/\ - 5x - 4 < 0.5 =» 3.5 < v 7 ! - 5x < 4.5 

=> 11.25 < -5x < 19.25 => -3.85 < x < -2.25. 
Step 2: |x - (-3)| <6 => -S < x + 3 <6 => -<5-3<x<<5-3. 

Then -6-3 = -3.85 => 6 = 0.85, or 6 - 3 = -2.25 => 0.75; thus 6 = 0.75. 



12.25 < 1 - 5x < 20.25 



10 

16 



W 

24 



-^>x>12or|<x<§. 



36. lim 4 - = i = 2 

x^2 x 2 

Step 1: || -2| <0.4 => -0.4 < | 

Step 2: |x-2|<<5^-<5<x-2<<5^.-<5 + 2<x<<5 + 2. 

Then -6 + 2 = f => <5 = |, or 6 + 2 = | => 6 = | ; thus «5 = 5 ■ 

37. Step 1: |(9-x)-5|<e=>-e<4-x<6=>-e-4<-x<e-4=>e + 4>x>4-e=>4-e<x<4 + e. 
Step 2: |x-4|<<5=^-<5<x-4<<5=^-<5 + 4<x<<5 + 4. 

Then -6 + 4 = -e + 4 =^ <5 = e, or<54-4 = e + 4 =>■ <5 = e. Thus choose <5 = e. 

38. Step 1: |(3x — 7) — 2| < e =>• -e < 3x - 9 < e =>■ 9 - e < 3x < 9 + e => 3-f<x<3 + f. 
Step 2: |x-3|<<5^-<5<x-3<<5^-<5 + 3<x<<5 + 3. 

Then -6 + 3 = 3 - f => <5=§,or<5 + 3 = 3+§ =>• § = f . Thus choose <5 = f . 



39. Step 1: k/x - 5 - 2 < e => -e < \fk - 5 - 2 < e => 2-e < a/x - 5 < 2 + e => (2 - e) 2 < x - 5 < (2 + e) 2 



=4> (2 - e) 2 + 5 < x < (2 + e) 2 + 5. 
Step 2: |x-9|<<5=^-6<x-9<<S=^-£ + 9<x<<S 
Then -<5 + 9 = e 2 - 4e + 9 ^ 5 = 4e - e 2 , or <5 + 9 
the smaller distance, S = Ae — e 2 . 



e 2 + 4e 4- 9 =$> 6 = 4e + e 2 . Thus choose 



40. Step 1: \j A - x - 2 < e => -e < ^4 - x - 2 < e => 2 - e < ^4 - x < 2 + e =^ (2 - e) 2 < 4 - x < (2 + e) 2 
=> -(2 + e) 2 < x - 4 < -(2 - e) 2 => -(2 + e) 2 + 4 < x < -(2 - e) 2 + 4. 



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Section 2.3 Precise Definition of a Limit 83 



Step 2: |x-0| <6 => -8 < x < 8. 



Then -S = -(2 + e) 2 + 4 = -e 2 - 4e =4> <5 = 4e + e 2 , or 5 = -(2 - e) 2 + 4 = 4e - e 2 . Thus choose 
the smaller distance, 8 = 4e — e 2 . 

H. Step 1: Forx^ 1, |x 2 - 1| < e => -6 < x 2 - 1 < e => 1 - e < x 2 < 1 + e => v/l ~ e < l x l < V 1 + e 
=>- V 1 — e < x < v 1 + e near x = 1. 
Step 2: |x - 1| < <5 =>- -<5 < x - 1 < <5 => -<5 + 1 < x < 6 + 1. 

Then -6 + 1 = sjl -e =>• 8 = 1- \fl - e, or 8 + 1 = y/l + e => 8 = yjl + e - 1. Choose 
8 = min < 1 — \J 1 — e, yl+e — 1 \ , that is, the smaller of the two distances. 



42. Step 1: For x ^ -2, |x 2 - 4| < e =>■ -e < x 2 - 4 < e => 4 - 6 < x 2 < 4 + e =^> ^4 - e < |x| < a/4 + e 
^> -i/4 + e<x<- a/4~ 



e near x = —2. 
Step 2: |x - (-2)| <S =>• -8 <x + 2<8 =>• -<5-2<x<<5-2. 



Then -6 - 2 = -a/4 + e =>• 6 = y/Z + e - 2, or <5 - 2 = -a/4- e => 5 = 2- a/4 - e. Choose 
<S = min { a/4+7 - 2, 2 - a/4 - e} . 



l-l|<e => -6 <i-Kc => l- e <i<l + e ^ T i 7 <x< T + 7 . 



43. Step 1: 

Step 2: |x-l|<<5=>-<5<x-l<<5^1-<5<x<l+<5. 



Thenl-^^ =» (5 = l- T i 7 = T ^,orl + 5 = T i 7 
Choose (5 = y+^< me smaller of the two distances. 



6= 1-^-1= r 5 -- 

1 — e 1 — e 



44. Step 1: 



1 _| ^- ,. -_> c <^ i i^-f^ 1 f ^ x ^ x 4- f __* l-3c ^ 1 ^ 1+3- . 3 . 2 ^ 3 

-J--|<£ _^ _ £< _ J __ <e ^ __ £< _ j< _ +£ =^ _____ < __ < _____ ^ _____ > X > j^gj 



TT37 < l x l < VT^' or 



TT37 <X< ^t^ for x near ^3. 
Step 2: |x- a/3| <5 =» -6<x- v / 3<6 => v / 3-<^<x<V / 3 + <5. 



Thenv^-« = VTT5I =► * = <fi " VOT or V^ + « = V T^C => « = V T^3_ " V^ 
Choose 8 = min -^ a/3 - 



3 / 3 

1 + 3f ' V 1 - 3e 



3 . 



45. Step 1: I (^f ) - (-6)1 <e => -e < (x - 3) + 6 < e, x ^ -3 =4> -e < x + 3 < e 

Step 2: |x - (-3)| <8 => -8 < x + 3 <8 => -<5 - 3 < x < <5 - 3. 

Then — 8 — 3 = — e — 3 => 8 = e, or 8 — 3 = e — 3 => 8 = e. Choose 8 = e. 

46. Step 1: I (^ff) - 2| < e =>■ -e < (x + 1) - 2 < e, x ^ 1 =4> l-e<x<l + e. 

Step 2: |x-l|<<5=>-<5<x-l<<5=^l-<5<x<l+<5. 

Then 1 — 8 = 1 — e => 8 = e, or 1 + 6 = 1 + e =$■ 8 = e. Choose 8 = e. 



=> -e-3<x<e-3. 



47. Step 1: x < 1: |(4 - 2x) - 2| < e => < 2 - 2x < e since x < 1. Thus, 1 - f < x < 0; 
x > 1: |(6x - 4) - 2| < e ^> 0<6x-6<e since x > 1. Thus, 1 < x < 1 + |. 
Step 2: |x-l|<<5^>-(5<x-l<<5=^l-6<x<l+<5. 

Then 1 — ^ = 1 — § ^> 5 = f , orl + <S=l + f ^ 5 = f . Choose 8 = f . 



48. Step 1: x < 0: |2x - 0| < e => -e < 2x < ^> - § < x < 0; 



x >0: 



< e => <x< 2e. 



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84 Chapter 2 Limits and Continuity 



Step 2: |x-0| <6 => -6 < x < 6. 



Then —6 



| =>. 5 = §, or <5 = 2e => 5 = 2e. Choose S = § . 



49. By the figure, — x < x sin 1 < x for all x > and — x > x sin - > x for x < 0. Since lim (— x) = lim x = 0, 

x x x^Ox^O 

then by the sandwich theorem, in either case, lim x sin - = 0. 

J x^ x 

50. By the figure, — x 2 < x 2 sin - < x 2 for all x except possibly at x = 0. Since lim (— x 2 ) = lim x 2 = 0, then 

by the sandwich theorem, lim x 2 sin - = 0. 

J x^O x 

51. As x approaches the value 0, the values of g(x) approach k. Thus for every number e > 0, there exists a S > 
such that < |x - 0| < 6 => |g(x) - k| < e. 



52. Write x = h + c. Then 0<|x-c|<<5<^>-<5<x-c<<5, x ^ c <J4> -<5 < (h + c) - c < <5, h + c^c 
■^ -S < h < S, h / ^ < |h - 0| < 6. 
Thus, limf(x) = L •£> for any e > 0, there exists 6 > such that |f(x) — L| < e whenever < |x — c| < 6 



^lf(h 



■ c) - Ll < e whenever < |h - 0| < 6 ^> limf(h + c) = L. 

h^0 



53. Letf(x) = x 2 . The function values do get closer to — 1 as x approaches 0, but lim f(x) = 0, not — 1 . The 

x — > 

function f(x) = x 2 never gets arbitrarily close to — 1 for x near 0. 

54. Let f(x) = sin x, L = |, and Xo = 0. There exists a value of x (namely, x = |) for which | sin x — 1 1 < e for any 

given e > 0. However, lim sin x = 0, not |. The wrong statement does not require x to be arbitrarily close to 

x — > z 

Xo. As another example, let g(x) = sin j, L = I, and Xo = 0. We can choose infinitely many values of x near 

such that sin - = \ as you can see from the accompanying figure. However, lim sin - fails to exist. The 

x i x — > x 

wrong statement does not require all values of x arbitrarily close to x = to lie within e > of L = | . Again 

you can see from the figure that there are also infinitely many values of x near such that sin - = 0. If we 

choose e < \ we cannot satisfy the inequality | sin 1 — \ | < e for all values of x sufficiently near Xo = 0. 
y 




55. A -91 <0.01 



-0.01 < 7Ti 



-9 < 0.01 



8.99 < ^f < 9.01 



^(8.99)<x 2 < 4(9.01) 



=$■ 2a /^ < x < 2i i^r or 3.384 < x < 3.387. To be safe, the left endpoint was rounded up and the right 
endpoint was rounded down. 



56. V = RI =>• ¥ =1 =>■ || -5| <0.1 => 

K IK I — 



-0.1 < 



120 



5 <0.1 



4.9 < l -f <5.1 



49 



R 

120 



51 



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Section 2.3 Precise Definition of a Limit 



85 



(120X10) < R < (120X10) 



23.53 < R < 24.48. 



51 - " - 49 

To be safe, the left endpoint was rounded up and the right endpoint was rounded down. 



x. Then |f(x) - 2| 



2-x>2-l = l. That is, 



x > 1. That is, 



57. (a) -6 < x - 1 < => 1 - 6 < x < 1 => f(x) 
jf(x) — 2 1 > 1 > i no matter how small S is ta 

(b) < x - 1 < 8 => 1 < x < 1 + 6 => f(x) = 
|f(x) — 1 1 > 1 no matter how small S is taken when 1 < x < 1 + <5 = 

(c) -6 < x - 1 < => 1 - 8 < x < 1 =>• f(x) = x. Then |f(x) - 1.5| = |x - 1.5| = 1.5 - x > 1.5 - 1 = 0.5. 
Also, 0<x-l<<5 => 1<x<1 + <5 => f(x) = x+l. Then |f(x) - 1.5| = |(x + 1) - 1.5| = |x - 0.5| 

= x — 0.5 > 1 — 0.5 = 0.5. Thus, no matter how small S is taken, there exists a value of x such that 
-6 < x - 1 < S but |f(x) - 1.51 > | =>• lim f(x) / 1.5. 

1 X — > 1 



ken when 1 - S < x < 1 =>■ lim f(x) ^ 2. 

X — > 1 

x+1. Then|f(x)-l| = |(x+l)-l| = |x| 
lim f(x)/l. 

X — > 1 



|h(x) - 4| = 2. Thus for e < 2, |h(x) - 4| > e whenever 2<x<2 + <5no 
=> lim h(x) ^ 4. 

x^ 2 ^ 

|h(x) - 3| = 1. Thus for e < 1, |h(x) - 3| > e whenever 2 < x < 2 + 8 no 
» lim h(x) ^ 3. 

x^ 2 ^ 

(c) For 2 - <5 < x < 2 =*• h(x) = x 2 so |h(x) - 2 1 - u - 
when x is near 2 and to the left on the real line =^ |x 



58. (a) For 2 < x < 2 + S =$> h(x) = 2 =; 
matter how small we choose 6 > 

(b) For 2 < x < 2 + 6 => h(x) = 2 =1 
matter how small we choose 6 > 



\- — 2 1 . No matter how small S > is chosen, x 2 is close to 4 
2 2| will be close to 2. Thus if e < 1, |h(x) - 2| > e 



whenever 2 — <5<x<2no mater how small we choose (5 > 



lim h(x) ^ 2. 

x^2 ^ 



59. (a) For 3 - S < x < 3 =^> f(x) > 4.8 => |f(x) - 4| > 0.8. Thus for e < 0.8, |f(x) - 4| > e whenever 



3 — <5<x<3no matter how small we choose 6 > 



lim f(x) / 4. 

x — > 3 



(b) For 3 < x < 3 



f(x) < 3 => |f(x) - 4.8| > 1.8. Thus for e < 1.8, |f(x) - 4.8| > e whenever 3 < x < 3 + S 



no matter how small we choose 6 > 



lim f(x) + 4.8. 

x — » 3 



(c) For 3 - 6 < x < 3 => f(x) > 4.8 => |f(x) - 3| > 1.8. Again, for e < 1.8, |f(x) - 3| > e whenever 3 - 6 < x < 3 



no matter how small we choose 6 > 



lim f(x) / 3. 

x — » 3 



60. (a) No matter how small we choose 6 > 0, for x near — 1 satisfying — 1 — 6 < x < — 1 + 6, the values of g(x) are 
near 1 => |g(x) — 2| is near 1. Then, for e = \ we have |g(x) — 2| > | for some x satisfying 

-1 -6 < x < -1 + <5,or0 < |x+ 11 < S => lim g(x) ^ 2. 

X — > — 1 

(b) Yes, lim g(x) = 1 because from the graph we can find a 6 > such that |g(x) — 1 1 < e if < |x — (— 1)| < 8, 

61-66. Example CAS commands (values of del may vary for a specified eps): 
Maple : 

f := x -> (x A 4-81)/(x-3);x0 := 3; 

plot( f (x), x=x0- 1 . .x0+ 1 , color=black, # (a) 

title="Section 2.3, #6 1(a)" ); 
L := limit( f(x), x=x0 ); # (b) 

epsilon := 0.2; # (c) 

plot( [f(x),L-epsilon,L+epsilon], x=x0-0.01..x0+0.01, 

color=black, linestyle=[l,3,3], title="Section 2.3, #6 1(c)" ); 
q := fsolve( abs( f(x)-L ) = epsilon, x=x0-l..x0+l ); # (d) 
delta := abs(xO-q); 

plot( [f(x),L-epsilon,L+epsilon], x=x0-delta..x0+delta, color=black, title="Section 2.3, #61(d)" ); 
for eps in [0.1, 0.005, 0.001 ] do # (e) 



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86 Chapter 2 Limits and Continuity 

q := fsolve( abs( f(x)-L ) = eps, x=xO-l..xO+l ); 

delta := abs(xO-q); 

head := sprintf("Section 2.3, #61(e)\n epsilon = %5f, delta = %5f\n", eps, delta ); 

print(plot( [f(x),L-eps,L+eps], x=xO-delta..xO+delta, 

color=black, linestyle=[l,3,3], title=head )); 
end do: 
Mathematica (assigned function and values for xO, eps and del may vary): 
Clear[f, x] 

yl: = L eps; y2: = L + eps; xO = 1; 
f[x_]: = (3x 2 - (7x + l)Sqrt[x] + 5)/(x - 1) 
Plot[f[x], {x, xO - 0.2, xO + 0.2}] 
L: = Limit [f[x], x -> xO] 
eps = 0.1; del = 0.2; 
Plot[{f[x], yl, y2},{x, xO - del, xO + del},PlotRange ->• {L - 2eps, L + 2eps}] 

2.4 ONE-SIDED LIMITS AND LIMITS AT INFINITY 



1 . (a) True 
(e) True 
(i) False 



(b) True 
(f) True 
(j) False 



(c) False 
(g) False 
(k) True 



(d) True 
(h) False 
(1) False 



(a) True 
(e) True 
(i) True 



(b) False 
(f) True 
(j) False 



(c) False 
(g) True 
(k) True 



(d) True 
(h) True 



(a) 


lim f(x) = | + 1 = 2, lim f(x) = 3-2 = 1 

x -» 2+ 2 x -> 2- 


(b) 


No, lim f(x) does not exist because lim f(x) ^ lim f(x) 

x ^ 2 W x ^ 2+ x ^ 2" W 


(c) 


lim f(x) = % + 1 = 3, lim f(x) = £ + 1 = 3 

x -> 4" - x -> 4+ £ 


(d) 


Yes, lim f(x) = 3 because 3 = lim f(x) = lim f(x) 

x ^ 4 x^4~ x ^ 4+ 


(a) 


lim f(x) = | = 1, lim f(x) = 3 - 2 = 1, f(2) = 2 

x -> 2+ l x -> 2- 


(b) 


Yes, lim f(x) = 1 because 1 = lim f(x) = lim f(x) 

x ^ 2 x ^ 2+ x ^ 2 


(c) 


lim f( x ) = 3-(-l) = 4, lim f(x) = 3 - (-1) = 4 

x — > -1 X — > -1 + 


(d) 


Yes, lim f(x) = 4 because 4 = lim f(x) = lim f(x) 

X— > -1 X -> -1 X -» -1+ 



5. (a) No, lim f(x) does not exist since sin Q) does not approach any single value as x approaches 



(b) lim f(x) 

x — > 



lim 

x->0- 



(c) lim f(x) does not exist because lim f(x) does not exist 

x -> x -» 0+ 

6. (a) Yes, lim g(x) = by the sandwich theorem since — */x < g(x) < ^/x when x > 

x — > 0+ 

(b) No, lim _ g(x) does not exist since \/x is not defined for x < 

x — > 

(c) No, lim g(x) does not exist since lim g(x) does not exist 

x — > x — > 



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Section 2.4 One-Sided Limits and Limits at Infinity 87 



7. (a) 



0, x=\ 



(b) lim f(x) 

X — ► 1 



1 



lim f(x) 

: — > 1+ 



(c) Yes, lim f(x) 
x — > 1 



1 since the right-hand and left-hand 
limits exist and equal 1 



8. (a) 




(b) lim f(x) = 

X — > 1 + 



lim f(x) 



(c) Yes, lim f(x) = since the right-hand and left-hand 

X — > 1 

limits exist and equal 



(a) domain: < x < 2 
range: < y < 1 and y = 2 

(b) lim f(x) exists for c belonging to 
(0,1)U(1,2) 

(c) x = 2 

(d) x = 



3 
f 

2 


y = 


it-x 2 , 0£x<l 

1, l<;t<2 

2, x = 2 

• 




"N 


\> • > 







1 2 



10. (a) domain: — oo < x < oo 
range: — 1 < y < 1 

(b) lim f(x) exists for c belonging to 

(-oo, -l)U(-l,l)U(l,oo) 

(c) none 

(d) none 




x, -1 <^x < or < x <_1 

^\ x = 

0, x < 1 or x > -1 



11. lim 

x^ -0.5" 



x + 2 
x- 1 



-0.5 + 2 
-0.5 + 1 



m = \/3 

1/2 V J 



12. lim J±=± 

x^i+ V x + 2 



l- l 

1 + 2 



^=0 



13- x jim 2+ (^r) (m = (=§£l) {Mjk) = <3 (I) 



14. 



l»n- (l+T) (^) (^) = (ih) W) W) = (*) (D (?) 



15. lim 

h->0+ 



Vh 2 + 4h + 5 - V5 



lim 

h-»0+ 



( 



y/h 2 + 4h + 5-y/E\ /y/h 2 + 4h + 5 + y/5 



\ / y/h 2 + 4h + 5+y/5 \ 

/ V Vh 2 + 4h + 5+x/5/ 



lim 



(h 2 + 4h + 5)-5 



lim 



h(h + 4) 



+ 4 



0+ h(Vh 2 +4h + 5+\/5) h-t0+ h(Vh 2 + 4h + 5 + v/5) 75 + V^ V^ 



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88 Chapter 2 Limits and Continuity 



16. lim 
h->0" 



lim 



v /6- x /5h2+llh + 6 _ ,. / y/e-x/Sh^+llh + e X / v/6+v/5h2+llh + 6 \ 

lf '" V h ) \j6+s/S&+Uh + 6) 

-h(5h + U) _ -(0+11) _ 



h->0 



6-(5h 2 +llh + 6) 



lim 



I) IF h I /6+\/5h2+llh + 6) h-»0 _ h(x/6 + V5h 2 + llh + 6) \/6+ v^ W6 

(|x + 2| = x + 2forx > -2) 



17. (a) Hm (x + 3) ^ = Km (x + 3) §±| 



-2+ 



-2+ 



lim (x + 3) = (-2) + 3=1 

x^ -2+ 



(b) x jim 2 _(x + 3)^ = 



Um_(x + 3)[=£ 



-(x + 2) 
2) 



(|x + 2| = -(x + 2)forx < -2) 



lim (x + 3)(-l) = -(-2 + 3) = -l 

x — > — 2 



18. (a) lim 



/2x(x-l) 



lim 



y/2x~(x-l) 



X-T1+ I-- !| x"l+ < X -D 

= lim v2x = v2 

x^ 1+ 

(b) lim 4^^ = lim ^ {x ~n 



|x - 11 = x - 1 forx > 1) 



|x- 1| = -(x- l)forx< 1) 



lim — V 2x = — v 2 
x — > 1 



19. (a) lim 



HI - 3 
3 



(b) e lim_ f = l 



20. (a) lim (t - Itl) =4-4 = 

t -> 4+ V L J ^ 



(b) t lim_(t-LtJ) = 4-3=1 



21. lim ^^ = lim ™* = 1 (where x = y^tf) 

6(^0 V2B x^O x V V7 



22. lim SS^J = lim ^^ 

t -» ' t -» o kt 



lim 

(9^0 



^ = k lim ^ 



23. lim ^ = i lim ^^ = ? lim ^ = | lim 



k- 1 =k 



sinfl _ 3 



■ o 4 y 



■ o 3 y 



o 3 y 4 e -> o 



(where 9 = kt) 
(where 9 = 3y) 



24. lim 
h-»0" 



sin 3h 



lim 
h-»0" 



'I . .J*-) - I ii m i 

,3 si „3hJ " 3 h ™ - (-») 



lim 

(9-0- 



1 



i (where (9 = 3h) 



25. lim Sa^ = hm liSS) = hm _Ms^_ = ( i im i ) ( i im 2_sm2x\ =1 . 2 = 2 

x^0 x x^0 x x^O xcos2x Vx^O co * 2x J Vx^O 2x / 



26. lim A = 2 lim -^ 

t^O tant t^Q (Mil) 



2 lim ^J = 2 ( lim cos ( 

t -> Sln ' Vt -> 



)W 



2-1-1 = 2 



27. lim ^^ = lim (-V 

x ¥ n cos 5x x ^ n V sin 2x cos 



i ) = (l i im * ") ( Hm i ) = (1-1) (1) = 1 

3 5x / V 2 x > q sin 2x y V x _;, q cos 5x / V 2 / v / 2 



28. lim 6x 2 (cot x)(csc 2x) = lim 6x2cosx = lim (3 cos x • -^ ^-) =3-1-1=3 

_ r, ■■ „ n Sin X sm 7x a V sin X Sin ZX / 



x^0 



x j, q sin x sin 2x x y n 



29. lim x + xcosx = lim (-^-* — + . xc ° sx = km (J L_) + hm -J- 

x ^ n sin x cos x x ^ n \ sin x cos x sin x cos x / x ^ n V sin x cos x / x y n sin x 



lim ' ! 

x^0 



x li- (^)+ x ^ ( + )=(1)(D+ 1=2 



30. lim 

x^0 2x 



lim , 

x^0 V2 



I + I(fc))=0-1 + !<1) = 



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Section 2.4 One-Sided Limits and Limits at Infinity 89 



31. lim 

t->o 



sin(l — cos t) 
1— cos t 



lim 

8^0 



1 since 9=1— cos t — > as t — > 



32. lim 

h->0 



sin (sin h) 
sin h 



lim 

0^0 



1 since 9 = sin h — » as h — > 



33. lim 42|j = lim 



sin 9 20\ _ 1 i- /sinfl 20 \ _ 1 i i _ 1 

sin 20 fl""*n V sin 20 '2d) 2 o J} n V sin 20 ) 2 ' i ' l 2 



6*^0 



34. lim ^ = lim (4^ • ^ • z) 

x j, n sin 4x x ► sin 



5 U (sin5x . _4x > 

4 y ► Q ^ ^x sin 4x / 



5.1-1=5 

4 A J- 4 



35. lim ^ = lim (^ - ^V) = lim (^ . i . *E . |> 

- sin fix Y , a V cos 3x sin 8x / Y , a V cos 3x sin 8x 3x 8 1 



x^O 



lim 



x^O 

1 \ / sin 3x \ / 8x 



x^O 



\ cos 3x 7 V 3x 7 V sin 8x7 



1-1-1 



36. lim sm3y "f 5y = lim sin 3y si " 4y c ° s 5y = lim ( s -^ 

y _> y cot 4y y _, g y cos 4y sin 5y y _, g ^ y 



\ ( sin 4y \ / cos5y \ / 3-4-5y \ 
^ ^cos4 y/ / \^sin5y,/ ^3-4-Sy^ 



y lim o (^)(^)( i || y )(^|)(M) = l. 1 .l. 1 .n = n 



Note: In these exercises we use the result lim -ij = whenever m > 0. This result follows immediately from 

v _^ 4- m x"V" n J 



± oo x 

Example 6 and the power rule in Theorem 8: lim ( -4 

F F X -» ± 00 v x 



lim (I) m/n =( lim iV 

X — > ± 00 X VX — > ± CO X / 



m/„ = Q_ 



37. (a) -3 



(b) -3 



38. (a) 7T 



(b) 7T 



39. (a) i 

40. (a) 1 



(b) § 



41. (a) -§ 



(b) 



42. (a) | 



(b) 1 



43. - i < 



1 ^ sin 2x ^ 1 



sin 2x 



lim 

X — > 00 x 



by the Sandwich Theorem 



44. 



1 £>■ COS ^ 1 

30 — 30 — 30 



lim 

e^ -c 



30 



by the Sandwich Theorem 



45. lim 2 r, t + si '" = lim 



-1+m 



t^oo t + cost t-»00 l + C^f 1 ) 



0-1+0 _ _i 
1+0 ' 



46. lim , ! + sh ". = lim - V ( !vL 

r — > oo 2r + 7-5sinr r _ > oc 2+^-5(^ 



li m i+o = I 

r l -H% 2 + 0-0 2 



47. (a) lim ^±| = lim |±| = f 



(b) | (same process as part (a)) 



48. (a) lim 



2x 3 + 7 



lim 



*oo x 3 — x 2 + x + 7 

(b) 2 (same process as part (a)) 



00 !-J + 7i + 73 



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90 Chapter 2 Limits and Continuity 



(a) lim 4±\ ..... , 

v/ x ^ oo x 2 + 3 x ^ oo 1 + 4 



1 , i 

lim_ ff£ = 

3 



(b) (same process as part (a)) 



50. (a) lim ^-^ = lim f^ 

x/ X — > OO x z - 2 x — > 00 1 — - 



(b) (same process as part (a)) 



51. (a) lim 



7x 3 



x — > oo x 3 - 3x 2 + 6x x — > oo 1 - I + 



lim 



7 (b) 7 (same process as part (a)) 



52. (a) lim 



x — > oo x 3 - 4x + 1 x 



lim 



1 _ 4 , 1 — U 



oo 1-^ + 



(b) (same process as part (a)) 



53. (a) lim 10x5 + f + 31 = lim 1±A±1 = «) 



X — > oo x u 



X — > oc 1 



(b) (same process as part (a)) 



9x 4 + > 



9 + 



54. (a) lim , 4 ^T 2 +X ,,- = lim T —, , — - - - 

v/ x — > oo 2x 4 + 5x 2 - x + 6 x ^ oo 2 + ^-4j + -^ 2 

(b) | (same process as part (a)) 



55. (a) lim ~ 2 /'rf + 3 = lim \ ? + fi = - \ 

w x-»x 3x J + 3x- ! -5i x — > oo 3 + - — ?j 3 



(b) — § (same process as part (a)) 



56. (a) lim —, — _ ., x _ , , « = lim - — , — k — g- = — 1 

v ' x — > oo x 4 - 7x 3 + 7x 2 + 9 x — > oo 1 - z 

(b) —1 (same process as part (a)) 



-l 



57. lim 



2,/x + x- 1 



lim 



x — > oo 3x-7 x — > oo 3 



58. lim 



lim 



x — » oo 2—Jx x — > oo ( 2 , i i 

v \7m) 



59. lim 



yx--x/x- 



lim 



l_ x (l/5)-(l/3) 



lim 



v-. 2 v - 



00 ^/x+^/x X — > -00 1 +x(V5)-(l/3) x _» _oc j j t 1 \ 



60. lim 



lim 



X^OCX 2 -x 3 x ^ oo 1 



oo 



6 1. lim 2xV 3 -xV 3 + 7 =Jim 2xV15 -^ + ^ 



x ™oo x 8 / 5 + 3x + y/x x — >"bo 1 + ^E + ^S~j 



OO 



,-n, ,- ?/x-5x + 3 .. 72/3 ~ 5 +x 

62. lim , v , 2/3 -, = lim f^ T 

x — > — oo 2x + x 2 / 3 -4 x — > — oo 2+-TH-- 



63. Yes. If lim f(x) = L = lim_ f(x), then lim f(x) = L. If lim f(x) ^ lim_ f(x), then lim f(x) does not exist. 

x » a~t~ x — > a x > a ^ ^ g+ x — > a. x > a. 

64. Since lim f(x) = L if and only if lim f(x) = L and lim_ f(x) = L, then lim f(x) can be found by calculating 

X > C Y i f*T X — > C X ► C 



lim f(x). 



65. If f is an odd function of x, then f(— x) = — f(x). Given lim f(x) = 3, then lim f(x) = — 3. 

x -> 0+ x -» 0- ' 



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Section 2.4 One-Sided Limits and Limits at Infinity 91 



66. If f is an even function of x, then f(— x) = f(x). Given lim f(x) = 7 then lim f(x) = 7. However, nothing 

x -> 2- x -> -2+ '6 

can be said about lim f(x) because we don't know lim f(x). 

x -> -2" x -> 2+ 



67. Yes. If lim ^ 

X — » oo g(x) 



2 then the ratio of the polynomials' leading coefficients is 2, so Jim 2gr 



2 as well. 



68. Yes, it can have a horizontal or oblique asymptote. 



69. At most 1 horizontal asymptote: If lim 



f(x) 
x — > i 6o g(x) 



L, then the ratio of the polynomials' leading coefficients is L, so 



lim 



fix) 



x — > -oo g(x) 



L as well. 



70. lim vx 2 + x — v x 2 — x = lim i/x 2 + x — y x 2 — x • 

X ^ OO v v x — > OO [ v 



\/x 2 + x + \/x 2 — X 



lim 



(x 2 + x) - (x 2 - x) 



\/x 2 + x + \/x 2 - x J X — > 00 ^x 2 + x + v/x 2 



lim 



2x 



V x 2 + x 4- \/x 2 — x X 



lim 



OO ./ill 



l + 7+\/l- 



1 ~~ 1 + 1 



71. For any e > 0, take N = 1. Then for all x > N we have that |f(x) - k| = |k - k| = < e. 

72. For any e > 0, take N = 1. Then for all y < -N we have that |f(x) - k| = |k - k| = < e. 



73. I = (5, 5 + 6) =4> 5 < x < 5 + 6. Also, ^\ - 5 < e => x - 5 < e 2 =4> x < 5 + e 2 . Choose <S = e 2 

=> lim \/ x ~ 5 = 0. 

x^5+ 

74. I = (4 - 6, 4) =>• 4 - 6 < x < 4. Also, ^/4 - x < e =>• 4 - x < e 2 =4> x > 4 - e 2 . Choose 6 = e 2 

=> lim_ \/4 - x = 0. 

x — > 4 

75. As x — > _ the number x is always negative. Thus, A — (— 1) < e =>• | ^ + l| < e =^> 0<e which is always 

true independent of the value of x. Hence we can choose any S > with — 6 < x < =>■ lim A = — 1. 

x — > l x l 



76. Since x — > 2 + we have x > 2 and |x — 21 = x — 2. Then, 



x-2 

|x-2| 



- 1 



x-2 
x-2 



1 <e=> 0< e 



which is always true so long as x > 2. Hence we can choose any 6 > 0, and thus 2 < x < 2 + 5 



x-2 

x— 21 



1 < e. Thus, lim A=| = 1 



x^ -2 



+ |x-2| 



77. (a) lim [ X J = 400. Just observe that if 400 < x < 401, then [xj = 400. Thus if we choose 6 = 1, we have for any 

x — > 400 + 

number e > that 400 < x < 400 + 6 => \ |xj - 400| = |400 - 400| = < e. 

(b) lim I x I = 399. Just observe that if 399 < x < 400 then |x|= 399. Thus if we choose S = 1, we have for any 

number e > that 400 - 6 < x < 400 => | LxJ - 399| = |399 - 399| = < e. 

(c) Since lim I x I ^ lim I x I we conclude that lim I x I does not exist. 

x _> 400+ L J x -» 400" L J x -> 400 L J 

78. (a) lim f(x) = lim ^/x = \/0 = 0; | y^x — 0| < e =^ -e < y/x < e => < x < e 2 for x positive. Choose 6 = e 2 

=> lim f(x) = 0. 

x^0+ 

(b) lim f(x) = lim x 2 sin (-) = by the sandwich theorem since — x 2 < x 2 sin (-) < x 2 for all x ^ 0. 

x^0 x^0 vx/ • vx/ 

Since |x 2 — 0| = j— x 2 — 0| = x 2 < e whenever |x| < yj~e, we choose 6 = yj~e and obtain |x 2 sin (ij — 0| < e 
if -6 < x < 0. 



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92 Chapter 2 Limits and Continuity 

(c) The function f has limit at Xq = since both the right-hand and left-hand limits exist and equal 0. 



79. lim xsini=lim \ sin 8 = 1, (8 = ±) 80. lim ^4= lim f 5 ^ = : 1, 



0^0 



-oo 1 + 1 g "q- ! + e ! 



i) 



81. lim 

X — > ± OO 



3x + 4 
2x-5 



lim 

x — > ± 00 



3 + ! 



lim |±| = | 

t-»0 2 ~ 5t 2 



(t=D 



82. lim ( i ) lA = lim z z = 1, (z = ±) 

J^M «' z ^ 0+ X 



83. lim (3 + |) (cos i) = lim (3 + 20)(cos 0) = (3)(1) = 3, (0 = 1) 

x^ ±00 x x 6* — » 

84. lim (4 -cos i) (l + sin 1 ) = lim (30 2 - cos 0) (1 + sin 0) = (0 - 1)(1 + 0) = -1, 

X — > OC ^ x X / V X / r\ q-[- 



2.5 INFINITE LIMITS AND VERTICAL ASYMPTOTES 



1. lim 

x^0+ 



OO 



(positive \ 
positive J 



2. lim 

x^O" 



2x 



-DC: 



positive \ 
negative J 



3. lim 

x^2" 



jc-2 



-oo 



(positive \ 
negative I 



4. lim — -; = oo 
x -> 3+ X ~ J 



( 



positive \ 
positive y 



5. lim 

x^ -8 



2x 
+ x+8 



-'00' 



(negative \ 
positive J 



6 - x Jj m 5 - 2x+T0 = °° 



(negative \ 
negative J 



1. 


lim , \. 2 


= oo 




9. 


(a) lim 

x - 0+ 


2 
3xV3 - 


oo 


10. 


(a) lim 

x -> 0+ 


2 
xV= - 


oo 


11. 


lim -Ar = 

x -» x ' 


= lim 

x^O 


4 

(xV5f 


13. 


lim tan x = oo 



(positive \ 
positive J 



8. lim 



x ^0 x '( x + 1 ) 



(b) lim 



Q- 3xV3 



(b) lim : 



-oo 



x^O" 



xV5 



12. lim -L = lim —I-, 

x^O x2/3 x^O (xV 3 ) 2 



14. lim sec x = oo 



(negative \ 
positive-positive J 



15. lim (1 + csc 6) = — oo 



16. lim (2 — cot 0) = — oo and lim (2 — cot 8) = oo, so the limit does not exist 

e -> o+ e^o- 



17. (a) x hm + x i L 4= x lim + ^^j 

(b) lim ^ = lim (x+2 x x * = ^ 



x _» 2" (x+2)(x-2) 
1 



(C) x i! m 2+ ^4= x _L im 2+ (x + 2)(x-2) 



(d) lim a 



lim 



2" * 2 -4 x -T-2 - (*+2)(x-2) 



f .. 1 .. ) 
\ positive-positive J 

( .. i ) 

V positive -negative / 

f .. i . ) 

^ positive-negative y 

r . i . 1 

V negative -negative / 



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Section 2.5 Infinite Limits and Vertical Asymptotes 93 



18. (a) lim 



lim 



x ""+ * 2 -l x "" + ( X + 1 K"-D 



oo 



(b) lim Jtx 

X — > 1 x ' 



lim , , * — rv 

x _» 1~ (x+l)(x-l) 

= lim 



-oo 



(c) lim -fa - ,„„ 

x^-l + x ' x— >-l + ( X+1 X X -!) 



(d) lim 



x^ -1- 



X 2 -l 



lim 



_ x - (x+l)(x-l) 



-OO 



19. 


(a) 


lim f - i = 

x^0+ 2 




(b) 


lim 4 ~ l = 

x -> 0" 2 x 




(c) 


lim £ - i 

s /^ 2 x 

x^ yi 




(d) 


lim £ - i = 

x--l 2 


20. 


(a) 


lim x2 -! - 

X TV 2x + 4 




(c) 


lim x2 -! = 

x" 1+ 2x + 4 




(d) 


lim " 2_1 = 
xTo- 2x + 4 


21. 


(a) 


lim + 4^W 

x _> 0+ x 2x 




(b) 


lim x V x + 2 2 

x _> 2+ x 2x 




(c) 


lim x2 3 " 3x , + 2 2 

x -> 2- x 2x 




(d) 


lim x2 -3 x + 2 . 

n X 3 - 2x 2 " 

x — > 2 x zx 



lim 

x^ 4 

lim 

x^ 0- 



1 



2- 3 
2 



1 

2 l/3 



-oo 

■ = oo 

2-V3 _ 2-1/3 



positive \ 



positive-positive J 

positive \ 
positive- negative J 

negative \ 



positive -negative 



negative 
negative -negative 



ative / 

M 

ltive y 



positive 



(i) 



3 



lim 



(x+l)(x-l) 
2x + 4 



positive \ 
positive y 

2-0 

2 + 4 



(b) 



ijm 2 _ f^ = -oo 



(e) lim 



o 



x 2 - 3x + 2 
x 3 — 2x 2 



lim 

.-»0 



(x-2)(x-l) 
f x 2 (x-2) 

(x-2)(x-l) 



x ^ 2 + x2 ( x " 2 > 



= lim 

x^2 

lim 

x^2 

lim 



(x-2)(x-l) 
x 2 (x - 2) 

(x-2)(x-l) _ 

x 2 (x - 2) 

(x-2)(x-l) 
x 2 (x - 2) 



= — oo 


/ negative-negative 
\ positive-negative 


= lim ^- = i x ^ 2 

x^2+ x2 4 ^ 




= lim i^i = i x ^ 2 

x^2- " 2 4 ' r 




lim M = 1 x ^ 2 

x -» 2 x2 4 ' r 




— OO 


/ negative-negative 



f positive \ 
I negative J 



x 2 - 3x + 2 



22 - ( a > lim 9+ ^^4x- 



(b) 
(c) 
(d) 

(c) 



2 

lim 



x 2 - 3x + 2 



lim 

x^0 



o+ x 3 — 4x 

2 -3x + 2 
x 3 -4x 



lim (x - 2)(x - 1) = 

._T2+ x ( x -2)( x + 2) 

lim < X - 2 >( X -D 

x _^_2+ x ( x " 2 X x + 2) 



lim 



(x-l) 



lim 



x 2 - 3x + 2 



(x-2)(x-l) 
5- x(x-2)(x + 2) 

lim fr- 2 **- 1 ) 

x Tl+ x ( x "2)(x + 2) 



lim .— 

x^l+ x3 - 4x 

lim x ~ \, = —00 

x-0+ x(x + 2) 

and lim x ~ \, = 00 

x — > _ x ( x + 2) 

so the function has no limit as x 



^2+ 


x(x + 2) 


2(4) 


lim (x 

x^-2+ x(x 


-1) _ 

+ 2) 


im 

->tr 


(x-l) 

x(x + 2) 


= OO 


. 


(x-l) 





-»1+ 


x(x + 2) 


(1)(3) 



OO 



0. 



(negative \ 
positive-positive J 

(negative 
negative-positive 



I negative \ 

I negative- positive J 

I negative \ 

I negative -positive J 



23. (a) lim \l 

t -> 0+ L 



-4-1 

tV 3 J 



(b) lim \2 



4-1 

tV 3 J 



00 



24. (a) lim [4j+7] 



t-»a 



(b) t lim_ [^+7] 



25. (a) lim [4 

x ^o+ L x/ 

(c) lim [-± 

x ^i+ L x2/ 



(x-l) 2 ' 3 
2 



x 2 /- 3 ' (x - 1) 2 / 3 



(d) x^l" [^ 



(x- l) 2 /' 3 
2 



x 2/3 T- ( X _ 1)2/3 



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94 Chapter 2 Limits and Continuity 



26. (a) lim [+, 
x _>0+ L xl/ 

(c) lim [^ 



1 



1/3 (x - 1)V3 

1 

1/3 (x- 1)4/3 



00 



27. y = -^ 

- 7 X — 



x-1 



(b) x^O- [^ 

(d) ^ [xi 



1 



xi/3 (x- I) 4 / 3 

1 
X l/3 (X- 1)4/3 



-OO 



-OO 



28- Y = d: 



x+l 





2 3 4 



X=- 1 




y = 



x + 1 



29- y = ^ 



2x + 4 











* 


\ 










10 














5 




1 

y 2* + 4 




.*" = 


-2 


v. 








-4 






2 - 


1 
-5 

-10 




1 2 



31. y = i±| = 1 + 4 

•' x + 2 x + 



+ 2 





-2 


'. \L5 


y= i±3 

_J_x + 2 


' **■ 




— . -5\ 

y= JT2 \ 


-2 








30. y = -4r 

J x — 3 




32. y 



2x q 2 

x+l — Z x+l 



1 ' 

2 A 1 
y " * + 1 / ,' 1 


2x 

y = 2 




/^ — " 


!/ 

1 

x = -1 J' 

j ' 


',-" 



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Section 2.5 Infinite Limits and Vertical Asymptotes 95 



33. y= -^ T=x+ i + -± T 



34. y 



x^+1 _ x+1 , 2 



x- 1 



x- 1 



_l — I — t_ 



"3 /yA> 

-2 




i' x 2 1 

X- 1 X- 1 



12 3 4 5 




35. y 



X— 1 X— 1 




36. y 



x 2 -l _ 1 
2x + 4 2 



X- 1 



2x + 4 




37. y 



£=i=x-i 



38. y 



x°+l 





39. Here is one possibility. 




40. Here is one possibility. 

y 




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96 Chapter 2 Limits and Continuity 
41. Here is one possibility. 



y=m 



42. Here is one possibility. 

y 




43. Here is one possibility. 



44. Here is one possibility. 



. 


, 




1 


5 








4 








3 


/W ~(*-2)= 


2 
1 









1 




3 4 5 





x - 3 



4 e 



45. Here is one possibility. 



46. Here is one possibility. 



-> 
j 
h(x) = — , x±0 

\x\ H 


; 

V 







>-l 





4 


f 


kw-i-sJt 






^— - — 


-4 -2 

-2 

-4 




/^ 4 6 



47. For every real number — B < 0, we must find a S > such that for all x, < |x — 0| < 6 =4> 4 < B. Now, 
- J J <-B<0o J j>B>0<^x 2 <i ■& Ixl < 4- . Choose 6 = 4- , then < Ixl < 6 => |x| < 4- 

x^ x^ B I I ,/b J-R I I ' ' VB 



-4 < -B so that lim - -4 

x2 x -> x 



-00. 



48. For every real number B > 0, we must find a 6 > such that for all x, < |x — 0| < 6 =>• A > B. Now, 
rr > B > O Ixl < i Choose 6 = i Then < |x - 01 < 6 => Ixl < ± => A > B so that lim A- = oo. 

|X| ■ ■ ±5 D ' ' ■ ■ ±5 | X| v > f) | x | 



For every real number B < 0, we must find a S > such that for all x, < |x — 3| < 6 



(x - 3)2 



< -B. 



Now, 



-2 

(x - 3)2 



< B < & ^^5 > B > & S-y^ < i <=> (x - 3) 2 < | ^ < \x - 3| < J I . Choose 



<5 = A /| , then < |x - 31 < 6 => — =% < -B < so that lim t-=|w 



50. For every real number B > 0, we must find a 6 > such that for all x, < |x — (— 5)| < 6 => ,„ , 1 C „ > B. 



(x + 5)2 



Now, 



(x + 5)2 



> B > O (x + 5) 2 < i O |x + 5| < 4j . Choose 5 = 4g . Then < |x - (-5)| < 6 



=> |x + 51 < 4- => 7— 4h > B so that lim 



^5 "^ (x+5)' 



x^ -5 



(x + 5)2 



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Section 2.5 Infinite Limits and Vertical Asymptotes 97 

51. (a) We say that f(x) approaches infinity as x approaches Xo from the left, and write lim_ f(x) = oo, if 

X — > x 

for every positive number B, there exists a corresponding number 6 > such that for all x, 
x - S < x < x => f(x) > B. 

(b) We say that f(x) approaches minus infinity as x approaches Xo from the right, and write lim f(x) = — oo, 

x^x+ 

if for every positive number B (or negative number — B) there exists a corresponding number S > such 
that for all x, x < x < x + 6 =^ f(x) < -B. 

(c) We say that f(x) approaches minus infinity as x approaches Xq from the left, and write lim_ f(x) = — oo, 

x —* x o 

if for every positive number B (or negative number — B) there exists a corresponding number S > such 
that for all x, x - 6 < x < x =4> f(x) < -B. 

52. For B > 0, - > B > <^> x < i Choose 6 = k. Then < x < <5 =4> 0< x < ^ => i > B so that lim i = oo. 

X D D D X ri-|- X 



53. For B > 0, \ < -B < O - ± > B > 4* -x < i & - i < x. Choose <5 = i Then -6 < x < 



s < x => - < -B so that lim - 

B x x^O" x 



-oo. 



54. For B > 0, ^ < -B ^ - ^ > B <=> -(x - 2)< A <^ x - 2 > - A «4> x > 2 - A, Choose § = A. Then 
2 - <5 < x < 2 =>• -<5 < x - 2 < => -i<x-2<0 =4> ^ < -B < so that lim_ ^A_ = ~°°- 

55. For B > 0, -^ > B <^> < x - 2 < ± Choose <5 = ± Then 2 < x < 2 + 6 =>• < x - 2 < 5 =4> < x - 2 < A 

=>- -A= > B > so that lim -A= = 00. 

x-z x^2+ X ~ Z 



56. For B > and < x < 1, y^Uj > B 44> 1 - x 2 < A O (1- x)(l +x) < A. Now A±* < 1 since x < 1. Choose 
<5 < i. Then 1 - <5 < x < 1 => -<5 < x - 1 < => 1 - x < <5 < ^ =^> (1- x)(l + x) < 



B V 2 J v B 



=> -r-^ > B for < x < 1 and x near 1 

1 — x z 

57. y = sec x + - 



lim 

i-»r 



00. 



58. y = sec x — \ 





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98 Chapter 2 Limits and Continuity 



59. y = tan x + i 



60. y = - — tan x 



y = tan x + 





61. y 



W- 



62. y 



1 

;j: = -2 

; 2 


1 ,1 


; V4-x 2 1 






1 2 
x=2 ; 



63. y = x 2 / 3 + 4, 




v*~- 




64. y = sin(^) 



y 


it 


_-^N 




LX 2 + 1 


-i 


i 





2.6 CONTINUITY 



1 . No, discontinuous at x = 2, not defined at x = 2 



2. No, discontinuous at x = 3, 1 = lim g(x) ^ g(3) =1.5 

x — > 3 



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Section 2.6 Continuity 99 



3. Continuous on [— 1, 3] 



4. No, discontinuous at x = 1, 1.5 = lim k(x) ^ lim k(x) = 

x -> 1 X -> 1+ 



5. (a) Yes 
(c) Yes 



(b) Yes, lim f(x) = 

X — > — 1 + 

(d) Yes 



6. (a) Yes, f(l) = 1 
(c) No 



(b) Yes, lim f(x) = 2 

X — > 1 

(d) No 



7. (a) No 



(b) No 



8. [-1,0)U(0,1)U(1,2)U(2,3) 

9. f(2) = 0, since lim f(x) = -2(2) + 4 = 0= lim f(x) 

x -> 2 x -> 2+ 

10. f( 1 ) should be changed to 2 = lim f(x) 

X — > 1 

11. Nonremovable discontinuity at x = 1 because lim f(x) fails to exist ( lim f(x) = 1 and lim f(x) = 0). 

x -> 1 x^l x^l + 

Removable discontinuity at x = by assigning the number lim f(x) = to be the value of f(0) rather than 



x^O 



f(0) = 1. 



12. Nonremovable discontinuity at x = 1 because lim f(x) fails to exist ( lim f(x) = 2 and lim f(x) = 1). 

X -> 1 X -» 1 X -> 1+ 

Removable discontinuity at x = 2 by assigning the number lim f(x) = 1 to be the value of f(2) rather than 

x — > 2 

f(2) = 2. 

13. Discontinuous only when x — 2 = => x = 2 14. Discontinuous only when (x + 2) 2 = => x = — : 

15. Discontinuous only when x 2 — 4x + 3 = =4> (x — 3)(x — 1) = =4> x = 3orx=l 

16. Discontinuous only when x 2 — 3x — 10 = => (x — 5)(x + 2) = =>■ x = 5 or x = — 2 

17. Continuous everywhere. ( |x — 1| + sin x defined for all x; limits exist and are equal to function values.) 

18. Continuous everywhere. ( |x| + 1 ^ for all x; limits exist and are equal to function values.) 

19. Discontinuous only at x = 

20. Discontinuous at odd integer multiples of |, i.e., x = (2n — 1) |, n an integer, but continuous at all other x. 

21. Discontinuous when 2x is an integer multiple of -it, i.e., 2x = rnr, n an integer => x= y,nan integer, but 
continuous at all other x. 

22. Discontinuous when y is an odd integer multiple of |, i.e., y = (2n — l) |, n an integer => x = 2n — 1, n an 
integer (i.e., x is an odd integer). Continuous everywhere else. 



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100 Chapter 2 Limits and Continuity 

23. Discontinuous at odd integer multiples of |, i.e., x = (2n — 1) |, n an integer, but continuous at all other x. 

24. Continuous everywhere since x 4 + 1 > 1 and — 1 < sin x < 1 =>• < sin 2 x < 1 => 1 + sin 2 x > 1 ; limits exist 
and are equal to the function values. 

25. Discontinuous when 2x + 3<0orx<— | =>• continuous on the interval [— |, oo) . 

26. Discontinuous when 3x— l<0orx<j =4> continuous on the interval [i, oo) . 

27. Continuous everywhere: (2x — l) 1 / 3 is defined for all x; limits exist and are equal to function values. 

28. Continuous everywhere: (2 — x) 1 / 5 is defined for all x; limits exist and are equal to function values. 

29. lim sin (x — sin x) = sin (it — sin it) = sin (it — 0) = sin it = 0, and function continuous at x = it. 

X — > w 

30. lim sin (| cos (tan t)) = sin (| cos (tan (0))) = sin (| cos (0)) = sin (|) = 1, and function continuous at t = 0. 

31. lim sec (y sec 2 y — tan 2 y — 1) = lim sec (y sec 2 y — sec 2 y) = lim sec ((y — l)sec 2 y) = sec((l — l)sec 2 1) 

y -> 1 y -> 1 " y -» 1 

= sec 0=1,, and function continuous at y = 1. 

32. lim tan [| cos (sin x 1 / 3 )] = tan [| cos (sin(0))] = tan (? cos (0)) = tan (|) = 1, and function continuous at x = 0. 



33. lim cos , - 

t— »o |_yi9-3sec2t 



i — cos -t— = cos ? = o > and function continuous at t = 0. 

V19-3SBC0J \/16 4 2 



34. lirn. -i/csc 2 x + 5v^3 tan x = Jcsc 2 (|) + 5\/3 tan (|) = w4 + 5^/3 (4- J = y^9 = 3, and function continuous at 



<;• 



35. g(x) = fff = (x + x 3Kx 3 ^ 3) = x + 3, x + 3 =► g(3) = x lim 3 (x + 3) = 6 

36. h(t) = t2 + 3t 2 1 ° = (t+ t 5 l (t 2 ~ 2) = t + 5, t ^ 2 =>■ h(2) = lim (t + 5) = 7 



37. f(s) 



-1 _ (s 2 + s+ l)(s- 1) _ sj+s + 1 



s 2 -! (s + l)(s-l) 



*£±Wl ^f(l) = s lim i (^±l) 



38. g(x) = -#^ = f + Tln = ^tt • x ¥= 4 =5- g(4) = lim (^±4) = | 

&v ' x 2 — 3x — 4 (x — 4)(x+l) x+1 ' * b\ i „ _^ a \x+l/ 5 

39. As defined, lim_ f(x) = (3) 2 — 1 = 8 and lim (2a)(3) = 6a. For f(x) to be continuous we must have 

x — > 3 x -> 3+ 

6a = 8 =^ a = \. 

40. As defined, lim g(x) = —2 and lim g(x) = b(— 2) 2 = 4b. For g(x) to be continuous we must have 
4b = -2 => b = -i 



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Section 2.6 Continuity 101 



41. The function can be extended: f(0) ~ 2.3. 



42. The function cannot be extended to be continuous at 
x = 0. If f(0) « 2.3, it will be continuous from the 
right. Or if f (0) w —2.3, it will be continuous from the 
left. 




-0.1 -0.05 



fix)' 



10 W - 1 



0.05 0.1 



43. The function cannot be extended to be continuous 
at x = 0. If f(0) = 1 , it will be continuous from 
the right. Or if f(0) = — 1 , it will be continuous 
from the left. 



44. The function can be extended: f(0) ~ 7.39. 



l(r 

0.5 



-0.1 -0.05 



-0.5 



i 



0.05 0.1 



/(*) = 



sin* 



-0.01 -0.005 



fix) = (1 + 2*)"* 




0.005 0.01 



45. f(x) is continuous on [0, 1] and f(0) < 0, f(l) > 
=^ by the Intermediate Value Theorem f(x) takes 
on every value between f(0) and f(l) => the 
equation f(x) = has at least one solution between 
x = and x = 1 . 




46. cos x 



(cos x) — x 



0. Ifx=-§, cos (-§)-(-§) 



> 0. If x 



COS I 



for some x between — | and | according to the Intermediate Value Theorem. 



— | < 0. Thus cos x — x = 



47. Let f(x) = x 3 - 15x + 1 which is continuous on [-4, 4]. Then f(-4) = -3, f(- 1) = 15, f(l) = - 13, and f(4) = 5. 
By the Intermediate Value Theorem, f(x) = for some x in each of the intervals —4 < x < — 1, — 1 < x < 1, and 
1 < x < 4. That is, x 3 — 15x +1=0 has three solutions in [—4, 4]. Since a polynomial of degree 3 can have at most 3 
solutions, these are the only solutions. 



Without loss of generality, assume that a < b. Then F(x) = (x — a) 2 (x - 
x, so it is continuous on the interval [a, b]. Moreover F(a) = a and F(b) 



b) 2 + x is continuous for all values of 
= b. By the Intermediate Value 



Theorem, since a < ^4^ < b, there is a number c between a and b such that F(x) 



Copyright (c) 2006 Pearson Education 




102 Chapter 2 Limits and Continuity 

49. Answers may vary. Note that f is continuous for every value of x. 

(a) f(0) = 10, f(l) = l 3 - 8(1) + 10 = 3. Since 3 < 7r < 10, by the Intermediate Value Theorem, there exists a c 
so that < c < 1 and f(c) = n. 

(b) f(0) = 10, f(-4) = (-4) 3 - 8(-4) + 10 = -22. Since -22 < -x/3 < 10, by the Intermediate Value 
Theorem, there exists a c so that —4 < c < and f(c) = — \J 3. 

(c) f(0) = 10, f(1000) = (1000) 3 - 8(1000) + 10 = 999,992,010. Since 10 < 5,000,000 < 999,992,010, by the 
Intermediate Value Theorem, there exists a c so that < c < 1000 and f(c) = 5,000,000. 

50. All five statements ask for the same information because of the intermediate value property of continuous 
functions. 

(a) A root of f(x) = x 3 — 3x — 1 is a point c where f(c) = 0. 

(b) The points where y = x 3 crosses y = 3x + 1 have the same y-coordinate, or y = x 3 = 3x + 1 

=> f(x) = x 3 - 3x - 1 = 0. 

(c) x 3 — 3x = 1 =4> x 3 — 3x — 1 = 0. The solutions to the equation are the roots of f(x) = x 3 — 3x — 1 . 

(d) The points where y = x 3 — 3x crosses y = 1 have common y-coordinates, or y = x 3 — 3x = 1 

=> f(x) = x 3 - 3x - 1 = 0. 

(e) The solutions of x 3 — 3x — 1 = are those points where f(x) = x 3 — 3x — 1 has value 0. 

51. Answers may vary. For example, f(x) = — _~ ' is discontinuous at x = 2 because it is not defined there. 
However, the discontinuity can be removed because f has a limit (namely 1) as x — > 2. 

52. Answers may vary. For example, g(x) = —j-t has a discontinuity at x = —1 because lim g(x) does not exist. 

( lim g(x) = — oo and lim g(x) = +oo. J 
\x->-l x ^_l+ / 

53. (a) Suppose Xo is rational =$■ f(xo) = 1. Choose e = \. For any 6 > there is an irrational number x (actually 

infinitely many) in the interval (xo — 6, Xo + S) =>■ f(x) = 0. Then < [x — Xoj < S but |f(x) — f(xo)| 
= 1 > I = e, so lim f(x) fails to exist =>- f is discontinuous at Xn rational. 

i X — > X 

On the other hand, Xo irrational =^ f(xo) = and there is a rational number x in (xo — 6, Xq + S) =^ f(x) 
= 1. Again lim f(x) fails to exist =>- f is discontinuous at Xn irrational. That is, f is discontinuous at 

X — > x 

every point, 
(b) f is neither right-continuous nor left-continuous at any point x because in every interval (x — 6, x ) or 
(xo, Xo + S) there exist both rational and irrational real numbers. Thus neither limits lim f(x) and 

x — > x 

lim f(x) exist by the same arguments used in part (a). 



54. Yes. Both f(x) = x and g(x) = x — | are continuous on [0, 1]. However -£i is undefined at x = | since 
g(i)=0 => -^ is discontinuous at x = 1. 

55. No. For instance, if f(x) = 0, g(x) = |~x] , then h(x) = ( |~x] ) = is continuous at x = and g(x) is not. 

56. Let f(x) = ^y and g(x) = x + 1. Both functions are continuous at x = 0. The composition f o g = f(g(x)) 

= , j._ l = i is discontinuous at x = 0, since it is not defined there. Theorem 10 requires that f(x) be 
continuous at g(0), which is not the case here since g(0) = 1 and f is undefined at 1. 

57. Yes, because of the Intermediate Value Theorem. If f(a) and f(b) did have different signs then f would have to 
equal zero at some point between a and b since f is continuous on [a. b]. 

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Section 2.7 Tangents and Derivatives 103 

58. Let f(x) be the new position of point x and let d(x) = f(x) — x. The displacement function d is negative if x is 
the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the 
Intermediate Value Theorem, d(x) = for some point in between. That is, f(x) = x for some point x, which is 
then in its original position. 

59. If f(0) = or f(l) = 1, we are done (i.e., c = or c = 1 in those cases). Then let f(0) = a > and f(l) = b < 1 
because < f(x) < 1. Define g(x) = f(x) — x =► g is continuous on [0, 1]. Moreover, g(0) = f(0) — = a > and 
g(l) = f(l) — 1 = b — 1<0 =4> by the Intermediate Value Theorem there is a number c in (0, 1) such that 

g (c) = => f(c) - c = or f(c) = c. 



60. Lete 



|f«0l 



> 0. Since f is continuous at x = c there is a 6 > such that |x — c| < 6 =4> |f(x) — f(c)| < e 



=> f(c) - e < f(x) < f(c) + e. 

If f(c) > 0, then e = \ f(c) => \ f(c) < f(x) < § f(c) =>■ f(x) > on the interval (c - 6, c + 6). 
If f(c) < 0, then e = - \ f(c) =^ § f(c) < f(x) < \ f(c) => f(x) < on the interval (c - 6, c + S). 




f(c>- 



C-<5 C+6 




f(c)+e 



61. By Exercises 52 in Section 2.3, we have lim f(x) = L <^> lim f(c + h) = L. 

x — > c h — > 

Thus, f(x) is continuous at x = c •<=>■ lim f(x) = f(c) <=> lim f(c + h) = f(c). 

x — > c h _> o 

62. By Exercise 61, it suffices to show that lim sin(c + h) = sin c and lim cos(c + h) = cos c. 

J h->0 v ' h-»0 v ' 

Now lim sin(c + h) = lim [(sin c)(cos h) + (cos c)(sin h)l =(sinc)( lim cos h ) + (cos c)( lim sin h) 
By Example 6 Section 2.2, lim cos h = 1 and lim sin h = 0. So lim sin(c + h) = sin c and thus f(x) = sin x is 
continuous at x = c. Similarly, 

lim cos(c + h) = lim [(cos c)(cos h) — (sin c)(sin h)l = (cos c) I lim cos h ) — (sin c) ( lim sin h ) = cos c. 
Thus, g(x) = cos x is continuous at x = c. 



63. x w 1.8794, -1.5321, -0.3473 



64. x sa 1.4516, -0.8547, 0.4030 



65. x w 1.7549 
67. x« 3.5156 
69. x« 0.7391 



66. xw 1.5596 



68. x w -3.9058, 3.8392, 0.0667 



70. xw -1.8955,0, 1.8955 



2.7 TANGENTS AND DERIVATIVES 



1. P i: mi = 1,P 2 : m 2 = 5 



2. Pi: mi = -2, P 2 : m 2 = 



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104 Chapter 2 Limits and Continuity 



3. Pi: mi = |,P 2 : m 2 



4. Pi: mi = 3, P 2 : m 2 = -3 



5. m = lim 



[4-(-l + h) 2 ]-(4-(-l) 2 ) 



lim 

h->0 



(l-2h + h 2 ) + l 



lim 
h->0 



h(2 - h) 



at (-1,3): y = 3 + 2(x-(-l)) => y = 2x + 5, 
tangent line 







y 

h 






y = 2x + 5 


/ C 








(-1,3)/ 


3 

2 
1 




V = 4-* 2 


-3 / 


/-2 -1 







I 2 \ 



6. m = lim " 1+h - 1)2+1 i-' (1 - 1)2+ " = lim 



h-' 



h->0 



h-»0 



lim h = 0; at (1, 1): y = 1 + 0(x - 1) =4> y 

h — > 



tangent line 



5 

\ * 

\ 3 


y.(x-1) 2 + 1 / 




(i.i) y-1 


-1 

-1 


1 2 



7. m = lim 
h->0 



2y/l+h-2yi 
h 

4(l+h)-4 



lim 

h^O 2h(%/l+h+l) 



lim 
h-»0 

lim 



2yi+h-2 2y/l+h + 2 
2vT+h + 2 

= 1; 



h-»0 Vl+h+l 
at (1,2): y = 2 + l(x - 1) =>• y = x + 1, tangent line 



, 


y = x + \/ 




4 






3 


A/y = l-{x 




2 
1 


- y<i,2) 







12 3 4 





J 1 



m= lim ±±±^ 

h->0 



i-(-i+hy 



lim — — r^^- 
h _> o h(-!+ h ) 2 



-(-2h + h 2 ) 



lim 

h ^0 h (-!+ h ) 2 



lim 



2-h 



h ^o (- 1 + h > 2 ' 
at (-1, 1): y = 1 + 2(x - (-1)) => y = 2x + 3, 
tangent line 




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Section 2.7 Tangents and Derivatives 105 



9. m = lim 
h->0 



(-2 + h) 3 -(-2) 3 



lim 

h->0 



-8+12h-6h 2 + h 3 + 8 



= lim (12-6h + h 2 ) = 12; 

IwO v ' 

at (-2, -8): y = -8 + 12(x - (-2)) 
tangent line 



y = 12x+ 16, 



y=12*+16 




10. m= lim '- 2+h]3 h '- 2 ' 3 
h->0 h 



-8-(-2 + h) 3 



h ™ -8h(-2 + h) 3 

j. -(12h-6h 2 + h 3 ) _ ,. 12-6h + tf 

hl^O -8h(-2 + h) 3 _ h lml 8(-2 + h) 3 



12 



_3_. 
16' 



at (-2, -I): y =-I_^(x-(-2)) 

=>■ y = — fg x — |, tangent line 



(-2,-1/8) 




16 2 



11. m = lim [P + ^ + H-s = lim (5 + 4h + hV5 = Um h(4 + h) = 4 

h -> h h -» h h -> h 

at (2, 5): y - 5 = 4(x - 2), tangent line 

12. m = lim 1(1 + h) - 2(1 h +h)2] - ( - 1) = lim il±!l=2=*=2*l±l = lim Wh) = _ 3 

h->0 h h->0 h IwO h 

at (1, —1): y + 1 = — 3(x — 1), tangent line 

13. m= lim 2±t 



3+h o 

13 + h) — 2 



h->0 



lim 



(3 + h)-3(h+l) 



lim 



-2h 



h — h(h+l) h — h(h+l) 



-2; 



at (3, 3): y — 3 = — 2(x — 3), tangent line 



14. m = lim ^^ = lim ^±^ = lim ^±g£^ = lim ^i+h) _ ^ 



h->0 



h ^* h(2 + h)2 h^o h(2 + h)2 _ h "o h (2 + h)2 - 4 



-2; 



at (2, 2): y - 2 = -2(x - 2) 



15. m= lim V±t^= lim (« + ^ + f + "3)-g = ^ h(i2 +6 h + tf) = 12 



h->0 



h->0 



h->0 



at (2, 8): y - 8 = 12(t - 2), tangent line 



16. m = lim ' (1+h)3 + 3 h (1+h) '- 4 = lim (' + 3h + 3h 2 + h 3 + 3 + 3h)-4 = ^ h(6 + 3h + h 2 ) = 

h->0 h h->0 h h->0 h 

at (1, 4): y - 4 = 6(t - 1), tangent line 



lim 



l 



n i- V4 + h-2 ,. V4 + h-2 V4 + 11 + 2 ,■ (4 + h)-4 

17. m = lim - — s = lim - — r • v , — = lim / , ; — ^ — "■" / , — \- — —r. — 

h->0 h h->0 h \/4 + h + 2 h^O h(v/4 + h + 2j h -> h(y4 + h + 2j v / 4 + 2 



|; at (4, 2): y — 2 = \ (x - 4), tangent line 



lim 



18. m = um v^+h)TT-3 = j. m ^+h-3 . ^+3 = lim (9 + h)-9 _ mii ____ 

h->0 h h-»0 h V9 + h + 3 h-»0 h(y9 + h + 3J h^O h( v /9 + h + 3j 

= 7^ = |; at (8,3): y - 3 = | (x - 8), tangent line 



19. Atx=-l,y = 5 =4> m = lim 



5(-l+h) 2 -5 
h 



lim 5( '- 2h h +h2) - 5 = lim 5h <4±W = -10, slope 
h^O h h-»0 h V 



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106 Chapter 2 Limits and Continuity 

20. Atx = 2,y = -3 => m= lim Ii^±^Mz5 = i im (i-4-4h-h 2 ) + 3 = Um zM 4 + h) = _ 4 j 

h->0 h h->0 h IwO h 



2 - (2 + h) 



21. At x = 3, y = i =4> m = lim ,3+h> -' 2 = lim tT^T^' = lim ,, , , , - 

' J 2 h^O h h^O 2h<2 + h) h -> 2h(2 + h) 4 



I, slope 



22. At x = 0, y = -1 => m = lim ' :+l . ( ' } = lim (h ~'^ + 1} = lim z^-r, = 2, slope 

J h-0 h h->0 h(h+1 » h->0 h(h+1) F 

23. At a horizontal tangent the slope m = => = m= lim K* + h > 2 + 4 <* + h >- i]-(x 2 + 4x- l) 

lim (* 2 + 2 *h + h 2 + 4x + 4h-l)-(x 2 + 4x-l) =; Hm (2xh + h !± 4h) = ^ (2x + h + 4) = 2x + 4; 



h-»0 



h->0 



h->0 



2x + 4 = => x= —2. Then f(— 2) = 4 — 8— 1 = — 5 =>• (—2, —5) is the point on the graph where there is a 
horizontal tangent. 



24. = m = lim 
h->0 



[(x + h) 3 - 3(x + h)] - (x 3 - 3x) __ ,. (x 3 + 3x 2 h + 3xh 2 + h 3 - 3x - 3h) - (x 3 - 3x) 



h->0 



= lim 3x 2 h + 3xh 2 + h 3 -3h = Hm / 3x 2 + 3 x h + h 2 - 3) = 3x 2 - 3; 3x 2 - 3 = =>■ x = -1 or x = 1. Then 
h->0 h IwO v ' 

f(— 1) = 2 and f(l) = — 2 =>• (—1,2) and (1, —2) are the points on the graph where a horizontal tangent exists. 



lim 



(x-l)-(x + h-l) 



lim 



-h 



1 



l i 

95 i — t-,-, — Htti <x+1 "~' i=- - ,,,,, - 

J ' hn o h ^"f, h(x-l)(x + h-l) h "^" h(x-l)(x + h-l) (x-1) 2 

v2 _ 1 _^ v 2 



=> (x - l) 2 = 1 => x 2 - 2x = => x(x - 2) = =>• x = or x = 2. If x = 0, then y = -1 andm = -1 
=>■ y=-l-(x-0) = -(x + 1). Ifx = 2, theny = 1 andm = -1 =4> y = 1 - (x - 2) = -(x - 3). 



r,r \ i- \Jx + h- \A l- \/x + h - ,/x v^ x + h + \/x ,. (x + h)-x 

26. 4 = m = lim c — — = hm c — — • ; , — ^ = lim 



h->0 



h^O 



/x + h+^/x h^O hK/x + h + 



nA) 



lim 



h^O h^Vx + h+yj) 2, A 

y = 2+i(x-4)=| + l 



^7= . Thus, I = -^-/= =>■ a/x = 2 =>• x = 4 =>• y = 2. The tangent line is 



2^/1 



2? lim f(2 + h)-f(2) = j im (100-4.9(2 + h) 2 )-(100-4.9(2) 2 ) = ^ -4.9 (4 + 4h + h 2 ) +4.9(4) 
' h^O h h->0 h h->0 h 

= lim (—19.6 — 4.9h) = — 19.6. The minus sign indicates the object is falling downward at a speed of 

h — > 

19.6 m/sec. 



28. lim nio + h)-f(io) = lim 3 ( io + h) 2 -3(io) 2 = Hm 3(201^) =60ft/sec _ 

h->0 h h->0 h h^O h 

29. lim '' ^^= lim " (3 + h) ! ~ " (3)2 = , lim, *P+6M*»-9] = lim ^ (6 + h) = 6 ^ 



h->0 



30. lim 

h->0 



h 




f(3 + h)- 


•f(3) 


h 




f(2 + h) - 


f(2) 



h^O 



lim 
h->0 



h->0 



h->0 



4 F(2 + h) 3 -fe(2) 3 _ f [12h + 6h 2 + h 3 ] _ 47r 



f(0 + h)-f(0) 



31. Slope at origin = lim 

v b h->0 h 

the origin with slope 0. 



lim 
h-»0 



lim 

h 2 sin(l) 



lim % fl2 + 6h + h 2 l = 16tt 
h^ 3 l J 



lim hsin(^) =0 =>• yes, f(x) does have a tangent at 



32. lim g( + I — — = lim T = lim sin ^ Since lim sin r does not exist, f(x) has no tangent at 



h->0 

the origin. 



h->0 



h->0 



h->0 



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33. lim 
h-»0~ 


f(o + h)-f(o )= Hm 

h twO" 


lim 

h->0 


f(0 + h)-f(0) =oo ^ 


34. lim 
h-»0~ 


u(o + h)-u(0 )= Hm 
h h-»0 



=t=fi = oo,and lim f(0 + h >- f(0) = lim I=° = 
h h->0+ h h ^0+ h 

oo =>• yes, the graph of f has a vertical tangent at the origin. 



Section 2.7 Tangents and Derivatives 

lim i -^- y = oo. Therefore, 



107 



V = oo,and lim u(0 + h > ~ u(0) 
h h->0+ h 



lim 
h-»0+ 



l- l 
h 



=>- no, the graph of f 



does not have a vertical tangent at (0, 1) because the limit does not exist. 



35. (a) The graph appears to have a cusp at x = 0. 




(b) lim 

h -> 0- 



f(0 + h)-f(0) 



lim 
h-»(r 



h 2 /=-0 



lim 



h 3/5 



-oo and lim 
h->0+ 



h 3 5 



limit does not exist 



the graph of y = x 2 / 5 does not have a vertical tangent at x = 0. 



36. (a) The graph appears to have a cusp at x = 0. 




(b ) Hm ffl + h)-f(0) = lim 
h -> 0" h h -> 0" 



h 4 / 5 - 



lim tjw 
h^0" hl/o 



— oo and lim 



lw0+ hV5 



oo =>- limit does not exist 



y = x 4 / 5 does not have a vertical tangent at x = 0. 



37. (a) The graph appears to have a vertical tangent at x = 0. 



(b) lim go + h)-f(0) = Hm hV^o = Hm 

h->0 h h->0 h h->0 



h 4 / 5 



(0,0) 



y = x 1 / 5 has a vertical tangent at x = 0. 



38. (a) The graph appears to have a vertical tangent at x = 0. 



(0,0) 



y =x 



3/5 



(b) lim f (° + h > f W = ij m h3/o ° = Hm -^ = 00^ the graph of y = x 3/5 has a vertical tangent 
w h-0 h h->0 h h-u bV ° & F J & 

at x = 0. 



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108 Chapter 2 Limits and Continuity 
39. (a) The graph appears to have a cusp at x = 0. 



y = 4x 2/5 -2x 




(b ) lim f(0 + h >- fW = lim 
h -> 0" h h -> 0" 



4h 2 / 5 - 2h 



lim 
h->0" 



h 3 .:, 



-co and lim 

h-»0+ 



h 3 .-. 



2 = oo 



=>• limit does not exist =4> the graph of y = 4x 2 / 5 — 2x does not have a vertical tangent at x = 0. 
40. (a) The graph appears to have a cusp at x = 0. 



5/3 - 2/3 
y ■ x - 5x 




(b) lim 
h->0 



f(0 + h) - f(0) 



lim 
h-»0 



h 5/3_ 5h 2/3 



lim h 2 / 3 



h->0 



7-yj = — lim j-^3 does not exist =4> the graph of 



y = x 5 / 3 — 5x 2 / 3 does not have a vertical tangent at x = 0. 

41. (a) The graph appears to have a vertical tangent at x = 1 
and a cusp at x = 0. 



y = x 2 / 3 -(x-l) 1 / 3 



(b) x = 1: lim d+h^-d+h- i)i/3-i 
h-»0 h 



lim 
h->0 



( 1 + h) 2 /' 3 - h 1 / 3 - 1 




y = x 2 / 3 — (x — l) 1 ' 3 has a vertical tangent at x = 1; 



x = 0: lim 
h-»0 



gO + h) - f(0) 



lim 
h->0 



h2/3-( h - 1)V3 -(-1)1/3 



lim 
h->0 



(h-1) 1 ' 3 , 1 



h 13 



does not exist =^ y = x 2//3 — (x — l) 1 / 3 does not have a vertical tangent at x = 0. 



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Section 2.7 Tangents and Derivatives 109 



42. (a) The graph appears to have vertical tangents at x = and 
x= 1. 



(b ) x = 0: lim f(0 + h '- f(0) 
h-0 h 



lim hl/3 + <h - 1 > 1/3 -^ 1 > 1/3 =oo 
h->0 h 



vertical tangent at x = 0; 



x = 1: lim 
h->0 



f(l+h)-f(l) 



lim 
h->0 

vertical tangent at x = 1. 



(1 + h) 1 / 3 + (l +h— l) 1 / 3 — 1 



'0.5 1.5 



y-x 1/3 *(x-l)l'3 



y = x 1 / 3 + (x - l) 1 / 3 has ; 



oo => y = x 1 / 3 + (x - l) 1 / 3 has a 



43. (a) The graph appears to have a vertical tangent at x = 0. 



(b) lim f(0 + h h ) - f(Q) 
W h->0+ h 



lim 
h-v(T 



f(0 + h)-f(0) 



lim 

lim 
h->0~ 



y/h-0 
h 

h 



lim -T- = oo; 

h->0 Vh 



lim =%$■ 



lim 

h->0~ 



=>- y has a vertical tangent at x = 0. 
44. (a) The graph appears to have a cusp at x = 4. 



(b ) lim f(4 + h >- f(4) 



lim 



lim 
h->CT 

=> y 



f(4 + h)-f(4) 



lim 
h->(T 



VK- 


-(4 + h)|- 


-0 




h 




VK- 


-(4 + h)| 






OO 




lim 

h^0+ 



lim -4- = oo; 
h^o+ v h 



lim 



h h -» o- h 

V 4 — x does not have a vertical tangent at x = 4. 



lim 

h-»tr 



45-48. Example CAS commands: 
Maple : 

f := x -> x A 3 + 2*x;x0 := 0; 

plot( f(x), x=x0- 1/2..X0+3, color=black, # part (a) 

title="Section 2.7, #45(a)" ); 
q := unapply( (f(x0+h)-f(x0))/h, h ); # part (b) 

L := limit( q(h), h=0 ); # part (c) 

secjines := seq( f(xO)+q(h)*(x-xO), h=1..3 ); # part (d) 

tanjine := f(xO) + L*(x-xO); 
plot( [f(x),tan_line, secjines], x=x0-l/2..x0+3, color=black, 



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110 Chapter 2 Limits and Continuity 

linestyle=[l,2,5,6,7], title=" Section 2.7, #45 (d)", 
legend=["y=f(x)", "Tangent line at x=0", "Secant line (h=l)", 
"Secant line (h=2)", "Secant line (h=3)"] ); 
Mathematica : (function and value for xO may change) 

Clear[f, m, x, h] 

x0 = p; 

f[x_]: = Cos[x] + 4Sin[2x] 

Plot[f[x],{x,x0- l,x0 + 3}] 

dq[h_]: = (f[x0+h] - f[x0])/h 

m = Limit[dq[h],h -> 0] 

ytan: = f[x0] + m(x - xO) 



yi 

y2 
y3 



f[x0] + dq[l](x - xO) 
f[x0] + dq[2](x - xO) 
f[x0] + dq[3](x - xO) 



Plot[{f[x],ytan,yl,y2,y3}, {x,xO - l,xO + 3}] 



CHAPTER 2 PRACTICE EXERCISES 



1. Atx 



f(x) 
lim f(x)=l=f(-l) 



1: lim 

x->-l 



lim f(x) 



x->-l 

f is continuous at x 



1. 



At x = 0: lim f(x) = lim f(x) = 
x -» 0" x -» 0+ 

Butf(0) = 1/ lim f(x) 

x — > 



lim f(x) = 0. 

x — » 



=> f is discontinuous at x = 0. 

If we define f(0) = 0, then the discontinuity at x = is 
removable. 



Atx = 1: lim f(x) 

X — > 1 



1 and lim f(x) = 1 

x^ 1+ 



lim f(x) does not exist 

x — > 1 

f is discontinuous at x = 1 . 



2. At x = -1: lim _ f(x) = and lim f(x) 

x — > -1 X -» -1+ 

=>• lim f(x) does not exist 

X — > —1 

=>• f is discontinuous at x = — 1 . 



Atx = 0: lim_f(x) 

x — > 



-oo and lim f(x) 
x^0+ 



=>- lim f(x) does not exist 
x — > 

=4> f is discontinuous at x = 0. 

Atx= 1: lim f(x) = lim f(x) -- 

x -> I" X -> 1+ 

Butf(l) = 0^ lim f(x) 

X — » 1 



lim f(x) = 1. 

X — > 1 



=4> f is discontinuous at x = 1 . 

If we define f(l) = 1, then the discontinuity at x = 1 is 
removable. 





A 
l<- 


y=/« 




1 

-1 




-1 




1 

1 


-*-x 



/(*) = 



l/x, < \x\ < 1 

0, * = 1 

1. x> I 


i 






1 -1 

V 


l 



3. (a) lim (3f(t)) = 3 lim f(t) = 3(-7) = -21 



(b) lim (f(t)r 
t -^ to 



( t lim f(t)) 2 = (-7) 2 = 49 



Copyright (c) 2006 Pearson Education 




Chapter 2 Practice Exercises 111 



(c) lim (f(t) • g(t)) = lim f(t) • lim g(t) = (-7)(0) = 

t — > tfl t — > to t — > to 



(d) lim 



t-»to 



f(0 

g(t)-7 



lim ft) 

t — »tQ 

lim (g(t) - 
t— *to 



lim f(t) 



lim o(t) — Hm7 
t-rto t->to 



-7 
0-7 



(e) lim cos (g(t)) = cos ( lim g(t) ) = cosO = 1 

t — » to ' \t — > t / 

'lim f(t)| = |-7|=7 



(f) lim |f(t)| = 

t — > to i i —i- K) l 

(g) lim (f(t) + g(t)) = lim f(t) 

t — * to t — > t 



lim g(t) 

t — » to 



(h) 



lim 

t->to 



(A) 



lim ft) 

t— »Q 



4. (a) lim -g(x) 

x — > 



- lim g(x) 
x — » 



(b) lim (g(x) • f(x)) = lim g(x) - lim f(x) = (y/l) (§} 
x^O x^O x^O V / 



(c) lim (f(x) + g(x)) = lim f(x) + lim g(x) 
x — > x^O x^O 

(d) lim ± - —^- - i - t 
V ' x^O f W 



lim f(x) 
x^o 



(e) lim (x + f(x)) = lim x + lim f(x) = - 



(f) Hm n 

x — > 



f(x)-cos X 
x-1 



lim f(x)- lim cos x 

x^O x^O 

lim x — lim 1 
x-tO x^O 



0-1 



5. Since lim x = we must have that lim (4 — g(x)) = 0. Otherwise, if lim (4 — g(x)) is a finite positive 

x^O x^O x^O 



number, we would have lim 
x->(r 



4-g(x) 



-oo and lim 

x^0 + 



4-g(x) 



oo so the limit could not equal 1 as 



x — ■> 0. Similar reasoning holds if lim (4 — g(x)) is a finite negative number. We conclude that lim g(x) = 4. 

x — > x — > 

6. 2 = lim x lim g(x) = lim x - lim lim g(x) = -4 lim lim g(x) = -4 lim g(x) 

x — > — 4 L x — > J x^-4 x ^ -4 Lx ^ J x-»-4 1x^1) J x — > 

(since lim g(x) is a constant) =>■ lim g(x) = ^ = — \ . 
x^O x^O 

7. (a) lim f(x) = lim x 1 / 3 = c 1 / 3 = f(c) for every real number c => f is continuous on ( — oo, oo). 

(b) lim g(x) = lim x 3 ' 4 = c 3 ' 4 = g(c) for every nonnegative real number c => g is continuous on [0, oo). 

(c) lim h(x) = lim x -2 / 3 = i = h(c) for every nonzero real number c =>• h is continuous on (— oo, 0) and (— oo, oo). 

(d) lim k(x) = lim x~ 1//(3 = i = k(c) for every positive real number c =^> k is continuous on (0, oo) 

8- ( a ) U (( n — I) 71 "; ( n + I) 71 ") ' wnere I = me set of all integers. 

nei 

(b) U ( n7r J ( n + l) 71 ")' where I = the set of all integers. 
nei 

(c) ( — oo, 7r) U (w, oo) 

(d) (-oo, 0) U (0, oo) 



(x - 2)(x - 2) 



lim 

x_>0- x ( x + 7 ' 

x 2 - 4x + 4 



oo and lim 

x^O 4 



(b) x hm 2 x 3 + 5x 2 _ 14x — ^^2 x(x + 7)(x-2) 



lim 



x(x + 7) 



(x-2)(x-2) 



lim 

-00 



x-2 



9. (a) ^lirr^ x 3 + 5x 2_ 14x - x lim o x(x + 7)(x _ 2 ) _ x "_^" x(x + 7) 



x^2; the limit does not exist because 



lim 



f ,~. ,x^2, and lim -^ 

x(x + 7) ' ' ' x 9 x(x 



x-2 





+ 7) 2(9) 



10. (a) lim * +1 3 

v ' x — > + + 

Now lim 



lim 



x(x+l) 



lim 



x+l 



-q x 3 (x 2 + 2x+l) x^To * a (*+l)(*+l) 



lim 



x 2 (x+l) 



,x/0 and x ^ — 1 . 



0- x2 ("+l) 



ooand x !IJV P ^+ T > 



lim 



x 5 + 2x 4 - x 3 



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112 Chapter 2 Limits and Continuity 



x(x+l) 



lim 



1 



(b) x lim i x 5 + X 2x 4 X +x 3 - x ^ m 1 X 3(x2 + 2x+l) - x ^ > "i 1 X 2 (x+1) 

1 



x / and x^-1. The limit does not 



exist because lim 



r . — — oo and lim .-,. 1 , , , = oo. 

j- x 2 (x+l) x _^ _ 1+ x 2 (x+l) 



11. lim V^ = lim 



l-x/x" 



lim 



x^l '-* x^l (l-V^)(l + v^) x^l 1 + \A 2 

12 ' x lml a xV^I = x lyjl a ( x 2 + a 2 )(x 2_ a 2) = x lim a ^2^^2 = 2? 



13. Hm (x + h) 2 -x 2 = lim ( x2 + 2hx + h 2 )-x 2 = Hm (2x + h) = 2x 

h->0 h h->0 h h->0 



14. lim 



(x + by - x' 



lim 

x^O 



(x 2 + 2hx + h 2 ) - x 2 



lim (2x + h) = h 

x — > 



_j l 

15 Hm 2±* i _ ,„„ _ ,,,,, 

x^O x x^O 2x(2 + x) x ^g 4 + 2x 



Um ^±4 = lim - 1 



16. Hm U±*Lll = Hm (x3 + 6x 2 + 12x + 8)-8 = ^ , 2 fe j 2) = 12 
x^O x x^O x x -> v 



17. lim [4 g(x)] 1 / 3 = 2 =>• [lim 4 g(x)l = 2 =>■ lim 4 g(x) = 8, since 2 3 = 8. Then lim g(x) = 2. 

x — > 0+ Lx — » 0+ J x — > 0+ x — > + 



18. lim xT ^- = 2 => lim (x + g(x)) = \ => y/S + lim g(x) = \ =► lim g(x) = \ - ^5 

Y 1 a / .^i a Y i -> / P\ Y V ■. / S Y 1 . / *! 



x^ x/5 



x^ V^ 



3x 2 + 1 



19. lim ^Y- = oo =>• lim g(x) = since lim (3x 2 + 1) = 4 



x^ 1 



x^ 1 



x -> 1 sOO 
20. lim ^™- = =4> lim g(x) = oo since lim (5 - x 2 ) = 1 



21. lim 



2x + 3 



lim 



2 + i _ 2 + _ 2 



x ^ oo 5x + 7 x^oos + i 5 + 5 



22. lim 



2x 2 + 3 



lim ^±1- 2 + o_2 



oo 5x 2 + 7 x — > -oo 5+4 5 + ° 5 



23. lim ^^ 

x — » -oo 3x J 



x ijm oc (i-^ + 3| 3 ) = o-o + o = o 



24. lim ^-4 



lim 



o 



x^ocx 2 -7x + l x^ool- 2 + 4, 1-0 + 



25. lim 



x"-7x 
00 x+1 



lim *=X 

— > —001 + 7 



-oc 



26. lim . 'o 4 ; ' . = lim 



x^oo 12x3 +128 x _^ 1 ii ool2 + l 2 8 



oc 



27. lim ^-r^ < lim r^ = since int x — > oo as x — » oo => lim ^r 11 ^ = 0. 



X^CO W — X — > oo L X J 



X — » 00 L X J 



28. lim ^^ < lim ^ = => lim ^^ = 0. 



29. lim 



x + sin x + 2 */x 



lim 



x ^ oo x + sinx x ^ oo 1 + 



"7^ _ 1+0+0 _ i 

* i + o x 



30. lim x2/3+x " 1 



lim 



x — > oo x z / J + cos^x x — > oo \ i 



1 + _ i 

1+0 



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Chapter 2 Practice Exercises 113 



31. Atx 



1: lim f(x) 

X — » — 1 



lim 



x(x 2 -l) 

1x2-1 



lim 



-1- 



x(x 2 -l) 

X 2-l 



lim _ x = — 1, and 



lim f(x) 



lim 



x(x 2 -l) 



lim 



x(x 2 -l) 

-(X2-1) 



1+ I* 2 - 'I X--1+ 

= lim (-x) = -(-1) = 1. Since 

X — > — 1 

lim _ f(x) ^ lim f(x) 

x —> -1 X —> -1+ 

=>• lim f(x) does not exist, the function f cannot be 

X — > — 1 

extended to a continuous function at x = — 1 . 




/C*)-;r(* , -l)/|* , -l| 



At x = 1 : lim f(x) = lim 

X — > 1 X — > 1 l x 'I 



lim (-x) 

X — > 1 



- 1 , and 



lim f(x) = lim ^4 — rp 



lim 



x(x 2 -l) 
x 2 -l 



lim x = 1 . Again lim f(x) does not exist so f 

; -> 1+ x-»l 



cannot be extended to a continuous function at x = 1 either. 



32. The discontinuity at x = of f(x) = sin (-) is nonremovable because lim sin - does not exist. 

\X/ X — > x 



33. Yes, f does have a continuous extension to a = 1: 



define f(l)= lim ^ 

x^ 1 x- 



34. Yes, g does have a continuous extension to a 



lim 



5 cosg 
40 - 2tt 



5 

4' 



35. From the graph we see that lim h(t) ^= lim h(t) 
6 v t -»• 0- r t -> 0+ 

so h cannot be extended to a continuous function 
at a = 0. 



X — ij. X 



9(8) 




5 cos 9 



h(t) 



*W = d + I/I) 1 ". fl = 



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114 Chapter 2 Limits and Continuity 



36. From the graph we see that lim k(x) ^ lim k(x) 

x -> x -> 0+ 

so k cannot be extended to a continuous function 
at a = 0. 



k x) 



k(x) = 



1 - 2W ' 



a = 



37. (a) f(— 1) = — 1 and f(2) = 5 => f has a root between — 1 and 2 by the Intermediate Value Theorem, 
(b), (c) root is 1.32471795724 

38. (a) f(— 2) = —2 and f(0) = 2 =>■ f has a root between —2 and by the Intermediate Value Theorem, 
(b), (c) root is -1.76929235424 

CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES 



1. (a) 



X 


0.1 


0.01 


0.001 


0.0001 


0.00001 


x x 


0.7943 


0.9550 


0.9931 


0.9991 


0.9999 



Apparently, lim x" 

x — » 0+ 



(b) 



0.6 



0.2 



y = x 



0.2 0.6 1 



2. (a) x 



l/(l„x) 



C) 1,,,n ' 



10 



100 



1000 



0.3679 0.3679 0.3679 

•i\VM 



Apparently, ^Hm, (±) A ° XJ = 0.3678 = 1 



(b) 



0.4. 



0.2 



f(x) 



J \ l/('» ») 



6) 



2 4 6 



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Chapter 2 Additional and Advanced Exercises 115 



3. lim_ L 

V — » c 



lim_ Ln 

V — > c 



1- 



lim v 2 



l - 



o 



The left-hand limit was needed because the function L is undefined if v > c (the rocket cannot move faster 
than the speed of light). 



1 < 0.2 



-0.2 < 



1 < 0.2 => 0.8 < Y < l - 2 =>■ L6 < V x < 2A => 2 - 56 < x < 5 - 76 - 



1 <0.1 



-0.1 < 



1 < 0.1 => 0.9 < Y < L1 => 1.8 < a/x < 2.2 => 3.24 < x < 4.84. 



5. 1 10 + (t - 70) x 10~ 4 - 10| < 0.0005 => |(t - 70) x 10~ 4 | < 0.0005 
=>■ -5 < t - 70 < 5 => 65° < t < 75° => Within 5° F. 



-0.0005 < (t - 70) x 10~ 4 < 0.0005 



1010 



6. We want to know in what interval to hold values of h to make V satisfy the inequality 
|V - 1000 1 = |367rh - 1000| < 10. To find out, we solve the inequality: 
|367rh - 1000| < 10 =^ -10 < 367rh - 1000 < 10 => 990 < 367rh < 1010 
=^> 8.8 < h < 8.9. where 8.8 was rounded up, to be safe, and 8.9 was rounded down, to be safe. 

The interval in which we should hold h is about 8.9 — 8.8 = 0.1 cm wide (1 mm). With stripes 1 mm wide, we can expect 
to measure a liter of water with an accuracy of 1%, which is more than enough accuracy for cooking. 



^ < h < 

367T — — 367T 



7. Show lim f(x) = lim (x 2 - 7) = -6 = f(l). 
x — > 1 x — > 1 



Step 1: |(x 2 - 7) + 6| < e => -e < x 2 - 1 < e =>■ 1 - e < x 2 < 1 + e => \J \ - t < x < a/I + e. 
Step 2: |x - 1| < 6 => -6 < x - 1 < 6 =$■ -6 + 1<x<6 + 1. 

Then -6+1= \/l - e or 6 + 1 = y/l + e. Choose 6 = min j 1 - \/ \ - e, y/l + e - l| , then 

< |x — ll < <5 =>• j(x 2 — 7) — 61 < e and lim f(x) = —6. By the continuity test, f(x) is continuous at x = 1. 

x — > 1 



Show lim i g(x) = lim ^ = 2 

4 



X^ | x 

Step 1: | i - 2| < e =4> -e < ^ - 2 < e =4> 2 - e < ^ < 2 + e 
Step 2: |x- j| < 6 => -6 < x- \ < 6 =4> -6+\<x<6+\ 

6 = — = — - — or 6 + - = — L - 

v 4 4+2e 4(2+e)' u1u ^4 4-2e 



Then -6 + \ 



4 + 2e 



4-2e ^ x ^ 4+2e ' 



Choose 6 = ,,„ £ , , , the smaller of the two values. Then < |x — t| < <5 => \w — 2| < e and lim ^- = 2. 

4(z+e) 14 1 I zx I i 2x 



4-2e 

I _]_ 
I 2x 



4(2 -0 



By the continuity test, g(x) is continuous at x 



9. Show lim h(x) = lim J2x - 3 = 1 = h(2). 

x^2 x^ 2 v 

Stepl: I y/2x - 3 - ll < e =S> -e < V2X -3-1 < e =^> l-e< ^2x - 3 < 1 + e =5> (1 ~f +3 <x< (1 + f + 3 . 



Step 2: |x - 2| < 6 => -6<x-2<6oi-6 + 2<x<6 + 2. 

Then -5 + 2 = ^#±^ => 6 = 2 - ( ^4 ±A = ^^ = e - £ , or 8 + 2 



(l+e) 2 + 3 
2 



s= (A±f+A_ 2 



(i + f) 2 -i 

2 



e + |- . Choose <5 = e — y, the smaller of the two values . Then, 



< |x — 2| < 6 => v 2x — 3 — 1 < e, so lim \J 2x — 3=1. By the continuity test, h(x) is continuous at x = 2. 



10. Show lim F(x) = lim ^9 - x = 2 = F(5). 

x — > 5 x — » 5 

Step 1: l-x/9 — x — 2| < e =4> -e < ^9 - x - 2 < e =4> 9 - (2 - e) 2 > x > 9 - (2 + e) 2 



Step 2: < |x - 5| < <*> =>• -<5 < x - 5 < <5 => -< < > + 5<x<<5 + 5. 



Then —(5 



9-(2 + e) 2 =► <S = (2 + e) 2 



2e, or * 



9 - (2 - e) 2 =>• 6 = 4 - (2 - e) 2 = e 2 



2e. 



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116 Chapter 2 Limits and Continuity 



Choose S 



2e, the smaller of the two values. Then, < |x — 51 < 6 



Vs 



2 < e, so 



lim v 9 — x = 2. By the continuity test, F(x) is continuous at x = 5. 

x — > 5 



11. Suppose Li and L 2 are two different limits. Without loss of generality assume L 2 > Li. Let e = I (L 2 — Li). 
Since lim f(x) = Li there is a <5i > such that < |x - x | < Si => |f(x) - Li| < e => -e < f(x) - Li < e 

X > Xq 

=> - i (L 2 - Li) + Li < f(x) < 1 (L 2 - Li) + Li => 4Li - L 2 < 3f(x) < 2Li + L 2 . Likewise, Hm f(x) = L 2 

so there is a 62 such that < |x — xo| < 82 => |f(x) — L 2 | < e => — e < f(x) — L 2 < e 
=► - i (L 2 - U) + L 2 < f(x) < i (L 2 - L : ) + L 2 => 2L 2 + L± < 3f(x) < 4L 2 - U 
=4> Li — 4L 2 < — 3f(x) < — 2L 2 — Lj. If 5 = min {61, 6 2 } both inequalities must hold for < |x — x | < S: 
4Lj - L 2 < 3f(x) < 2Lj 

U - 4L 2 < -3f(x) < -2L 2 

a contradiction. 



a + u \ 

L 2 - Li J 



=> 5(Li - L 2 ) < < L x - L 2 . That is, Li - L? < and U - L 2 > 0, 



12. Suppose lim f(x) = L. If k = 0, then lim kf(x) = lim = = 0- lim f(x) and we are done. 

cr X — > C X— > C X — > C X — > c 

If k ^ 0, then given any e > 0, there is a 6 > so that < |x - c| < 6 =>• |f(x) - L| < j|r =4> |k||f(x) - L| < e 
=$► |k(f(x) - L)| < e => |(kf(x)) - (kL)| < e. Thus, lim kf(x) = kL = k( lim f(x)Y 

13. (a) Since x — > + , < x 3 < x < 1 =4> (x 3 - x) — ► 0" =4> lim f (x 3 - x) = lim f(y) = B where y = x 3 - x. 

v ; x-o+ y-o- : ' J 

(b) Since x — *• 0~, -1 < x < x 3 < => (x 3 - x) -^ 0+ =>• lim f (x 3 - x) = lim f(y) = A where y = x 3 - x. 

V ' X^O" V ' y ^0+ 

(c) Since x -*• 0+, < x 4 < x 2 < 1 =>■ (x 2 - x 4 ) -> 0+ => lim f(x 2 -x 4 )= lim f(y) = A where y = x 2 - x 4 . 

x — » + y — > 0+ 

(d) Since x -> 0~, -1 < x < => < x 4 < x 2 < 1 => (x 2 - x 4 ) -> 0+ =>■ lim f (x 2 - x 4 ) = A as in part (c). 

x — > 0+ 



14. (a) True, because if lim (f(x) + g(x)) exists then lim (f(x) + g(x)) — lim f(x) = lim [(f(x) + g(x)) — f(x)] 

x — ► 2l x — ^ 3. x — y a x — * a 

= lim g(x) exists, contrary to assumption. 

(b) False; for example take f(x) = 1 and g(x) = — -. Then neither lim f(x) nor lim g(x) exists, but 

x ' x x^O x — > 

lim (f(x) + g(x)) = lim (I - i) = lim = exists, 
x — > x ^ x x x^O 

(c) True, because g(x) = |x| is continuous => g(f(x)) = |f(x)| is continuous (it is the composite of continuous 
functions). 

x < 



(d) False; for example let f(x) = < ' 
continuous at x = 0. 



x> 



f(x) is discontinuous at x = 0. However |f(x)| = 1 is 



x z -l 



15. Show lim f(x) = lim , 

x — > — 1 x —> -1 x + ' 



x->-l ( x+1 > 
Define the continuous extension of f(x) as F(x) = < 



l im (» + ix» : D = _2, x ^-1, 



x -1 x ■+ _ 1 

x + 1 ' ' . We now prove the limit of f(x) as x — » 1 

-2 , x = — 1 



exists and has the correct value. 



Step 1: 



x+l 



-(-2) < e => -e< 



(x+l)(x-l) 
(x+l) 



2 < e => -e < (x - 1) + 2 < e, x^-1 =>• -e-l<x<e-l. 



Step 2: |x - (-1)| < (5 => -6 < x + 1 < S => -S-l<x<S-l. 

Then -6 - 1 = -e - 1 => S = e,OT 8 - I = e - 1 => 6 = e. Choose S = e. Then < |x - (-1)| < S 



\x l -l 



(—2) < e =$■ lim F(x) = —2. Since the conditions of the continuity test are met by F(x), then f(x) has a 



I x+1 'I x-> -1 

continuous extension to F(x) at x = — 1 



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Chapter 2 Additional and Advanced Exercises 117 



16. Show lim g(x) = lim 
x — > 3 x — > 3 



x 2 - 2x - 3 
2x-6 



lim 

x^ 3 



(x-3)(x + l) 



Define the continuous extension of g(x) as G(x) 



2(x-3) - 2, X 7^ 3. 

r x^-2x-3 x ^ 3 



3 exists and has the correct value. 



x 2 -2x-3 



2 < e => -e < 



(x - 3)(x + 1) 



Step 1: | 2x6 ^| ^ ^ -, >, - 2(x-3) 

Step 2: |x - 3| < 6 => -6<x-3<6 =>• 3-6 <x<6 



2x - 6 ' ' . We now prove the limit of g(x) as 



2< e => _ e< *±!_ 2 <e, x ^ 3 => 3 - 2e < x < 3 + 2e. 



Then, 3 - 6 = 3 - 2e 

I x 2 - 2x - 3 



2x-6 



2 < e =► lim ( *- ( f ^ 

x^3 2(x-3) 



2e, or 6 + 3 = 3 + 2e => 5 = 2e. Choose <5 = 2e. Then < |x - 3| < 6 
2. Since the conditions of the continuity test hold for G(x), 



g(x) can be continuously extended to G(x) at x = 3. 

17. (a) Let e > be given. If x is rational, then f(x) = x => |f(x) — Oj = |x — 0| < e o- |x — 0| < e; i.e., choose 
6 = e. Then |x — 0| < 6 =4> |f(x) - 0| < e for x rational. If x is irrational, then f(x) = => |f(x) - 0| < e 
■& < e which is true no matter how close irrational x is to 0, so again we can choose 6 = e. In either case, 
given e > there is a 6 = e > such that < |x — 0| < <5 =>■ |f(x) — 0| < e. Therefore, f is continuous at 
x = 0. 
(b) Choose x = c > 0. Then within any interval (c — 6, c + 6) there are both rational and irrational numbers. 
If c is rational, pick e = |. No matter how small we choose 6 > there is an irrational number x in 
(c — 6, c + 6) =^ |f(x) — f(c)| = |0 — c| = c > | = e. That is, f is not continuous at any rational c > 0. On 
the other hand, suppose c is irrational => f(c) = 0. Again pick e = | . No matter how small we choose 6 > 

3c -. Then |f(x) - f(c)| = |x - 0| 



e ■& | < x < 



there is a rational number x in (c — 6, c + 6) with |x — c| < | = 
= |x| > | = £ =>• f is not continuous at any irrational c > 0. 

If x = c < 0, repeat the argument picking e = '-y = ^ ■ Therefore f fails to be continuous at any 
nonzero value x = c. 



18. (a) Let c = ™ be a rational number in [0, 1] reduced to lowest terms =^> f(c) = j. Pick e = ^. No matter how 
small 6 > is taken, there is an irrational number x in the interval (c — 6. c + 6) =>■ |f(x) — f(c)| = |0 — - 1 



- > tj- = e. Therefore f is discontinuous at x = c, a rational number. 

n 2n ' 



(b) Now suppose c is an irrational number =>• f(c) = 0. Let e > be given. Notice that | is the only rational 
number reduced to lowest terms with denominator 2 and belonging to [0, 1]; I and | the only rationals with 
denominator 3 belonging to [0, 1]; \ and | with denominator 4 in [0, 1]; \, |, I and I with denominator 5 in 
[0, 1]; etc. In general, choose N so that i < e =^> there exist only finitely many rationals in [0, 1] having 
denominator < N, say ri, r2, ..., r p . Let 6 = min{|c — r ; | : i=l,...,p}. Then the interval (c — 6. c + 6) 
contains no rational numbers with denominator < N. Thus, < |x — c| < 6 =^> |f(x) — f(c)| = |f(x) — 0| 
= |f(x)| < i < e => f is continuous at x = c irrational. 



Copyright (c) 2006 Pearson Education 




118 Chapter 2 Limits and Continuity 



(c) The graph looks like the markings on a typical ruler 
when the points (x, f(x)) on the graph of f(x) are 
connected to the x-axis with vertical lines. 



0.8 



0.6 



0.4 



0.2" 



m. 



im 



m 



0.2 



0.4 



0.6 



0.8 



f( , = f 1 /" itx = m/ni 
/w [0 if x is irratio; 



is a rational number in lowest terms 
irrational 



19. Yes. Let R be the radius of the equator (earth) and suppose at a fixed instant of time we label noon as the 
zero point, 0, on the equator =4> + irR represents the midnight point (at the same exact time). Suppose Xj 
is a point on the equator "just after" noon =4> Xj + irR is simultaneously "just after" midnight. It seems 
reasonable that the temperature T at a point just after noon is hotter than it would be at the diametrically 
opposite point just after midnight: That is, T(xi) — T(xi + 7rR) > 0. At exactly the same moment in time 
pick x-2 to be a point just before midnight => X2 + irR is just before noon. Then T(X2) — T(X2 + ttR) < 0. 
Assuming the temperature function T is continuous along the equator (which is reasonable), the Intermediate 
Value Theorem says there is a point c between (noon) and ttR (simultaneously midnight) such that 
T(c) — T(c + 7rR) = 0; i.e., there is always a pair of antipodal points on the earth's equator where the 
temperatures are the same. 



20. 



x lim f(x)g(x) = x lim 1 [(f(x) + g(x)) 2 - (f(x) - g(x)) 2 ] = \ [( x lim (f(x) + g(x)))' - ( x lim(f(x) - g(x)) 



H3 2 -(-ir 



21. (a) Atx = 0: lim r + (a) = lim 

a — > a — > 



-1 + Jl + i 



lim 

a^0 



lim 



l-(l+a) 



a ^0 af-1-Vl + a) -l-yi + O 



At x = — 1: lim r + (a) = lim 

a — > — 1 + a — 

(b) At x = 0: lim r_ (a) = lim 

a — > a — » 



l-(l + a) 



! )fe£§) 



lim 



1+ a(-l-Vl + a) a->-l a (-1-^/1+ a) -1 - v^ 

1- x/TTe 



a^0" V a / \-l + s/T+IJ 



lim 



t-q + a) 



lim 



lim 



-l 



a^0" a(-l + Vl + a) a -> 0" a (-1 + ^1 + a) a _, Q" -1 + Vl + a 



oo (because the 



denominator is always negative); lim r_(a) = lim -, = — oo (because the denominator 



a^0+ - 1 + 1 



is always positive). Therefore, lim r_ (a) does not exist. 

a — > 



At x = — 1: lim r_(a) 
a->-l+ 



lim 



-1- Vl + a 



lim 4— = 1 

_> _1+ -l + V!+ a 



Copyright (c) 2006 Pearson Education 




Chapter 2 Additional and Advanced Exercises 119 



(c) 




(d) 



-0.5 0.5 1 

Graph not to scale 




r (a) 



a=0.05 



1 1 






/ 


r-(o)=- 


-\--JTTa 
a 


1 

-2 


2 ^_4___ 


S- 8 


-4 







a=0.2 




22. f(x) = x + 2 cos x =>• f(0) = + 2 cos = 2 > and f(-7r) = ~n + 2 cos (-tt) = -n - 2 < 0. Since f(x) is 
continuous on [— ir, 0], by the Intermediate Value Theorem, f(x) must take on every value between [—tt — 2, 2]. 
Thus there is some number c in [— tt, 0] such that f(c) = 0; i.e., c is a solution to x + 2 cos x = 0. 



23. (a) The function f is bounded on D if f(x) > M and f(x) < N for all x in D. This means M < f(x) < N for all x 
in D. Choose B to be max {|M| , |N|} . Then |f(x)| < B. On the other hand, if |f(x)| < B, then 

B < f(x) < B => f(x) > -B and f(x) < B =>■ f(x) is bounded on D with N = B an upper bound and 
M = — B a lower bound. 



(b) Assume f(x) < N for all x and that L > N. Let e 



L-N 



Since lim f(x) = L there is a 6 > such that 



xn 



< |x - x | < 6 => |f(x) - L| < e <S> L - e < f(x)< L + e <S> L - ^ < f(x)< L + ^ 



O ^±S < f(x) < 



3L-N 



But L > N 



L + N 



f(x) < N. This contradiction proves L < N. 



> N => N < f(x) contrary to the boundedness assumption 



(c) Assume M < f(x) for all x and that L < M. Let e = ^y^. As in part (b), < |x - x j < 6 



< f(x) < L + 



<£• 



3L-M 



< f(x) < 



< M, a contradiction. 



24. (a) If a > b, then a - b > => |a - b| = a - b => max (a, b) = ^ 



i + b I a — b _ 

2 ' 2 — 2 



If a < b, then a - b < =>• 
= ^ =b 

2 u - 

(b) Let min (a, b) - ^ - ^^ 



(a — b) = b — a =4> max (a, b) 



a + b 
2 



|a-b| 



2a 
2 

a + b 
2 



b-a 
2 



a + b 

2 



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120 Chapter 2 Limits and Continuity 



25. lim 



sin(l — cos x) 



sin(l-cosx) 1 -cosx 1 + cos x 



: lim 



lim 



sin(l — cos x) 



lim )r cos2x , = 1 ■ lim 



1 + cos x x j, q 1 — cos x x ^ q x(l + cos x) x j, q x(l + cos x) 



lim ^ . ^mjL_ =1 . (2) =0 . 

x _^ q x 1 + cos x V 2 / 



26. lim -^- = lim ^ • -^ %. 

x _» o+ sm V x x -» + x Sln V x V x 



1 • lim t-^y • lim v/x = 1 • • = 0. 



27. lim ™(^) = lim *M^A . m* = i im ^i^l . lim ^ = , . i = L 



x^O 



x^O 



sin x x 



x^O 



x^O 



28. lim 



sin(x 2 + x) _ j- m sin(x 2 +x) . ( v j_ i\ _ ]; m sin(x 2 + x) 
x^O x x ^o x2 + x 



(x + 1) = lim =J^2 • lim (x + 1) = 1 • 1 = 1 

V ; x^O x2 + x x^0 V 



29. lim 

x^2 



sin(x 2 - 4) _ ,. sin(x 2 - 4) 



sin(x 2 — 4) 



, - lim ^f^l . ( x + 2) = lim sn ^=^ • lim (x + 2) = 1 • 4 = 4 

x - 2 x^2 x2 - 4 V ; x^2 x2 - 4 x^2 V ' 



30. lim ^(^-3) = lim Mf-z) . _^ = lim Mf-z) . , im 
x^9 x ~ 9 



- = 1 • - = - 
x't^'g v^- 3 \/ x + 3 x-T9 \A- 3 x i ^' 1 9\/ x + 3 6 6 



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CHAPTER 3 DIFFERENTIATION 



3.1 THE DERIVATIVE OF A FUNCTION 



1. Stepl: f(x) = 4-x 2 andf(x + h) = 4-(x + h) 2 

Stet) 2' f ( x + h >- f W = K-(x + h) 2 ]-(4-x 2 ) _ (4-x 2 -2xh-h 2 )-4 + x 2 _ -2xh-h 2 _ h(-2x-h) 
" ' h h h h h 

= -2x - h 

Step 3: f (x) = lim (-2x - h) = -2x; f (-3) = 6, f (0) = 0, f (1) = -2 

h — > 



2. F(x) = (x - l) 2 + 1 and F(x + h) = (x + h - l) 2 + 1 => F'(x) = lim 



[(x + h-l) 2 + l]-[(x-l) 2 +l] 



h-»0 



,■ (x 2 + 2xh + h 2 -2x-2h+l + l)-(x 2 -2x+l + l) _ 

h™0 h 

2(x - 1); F'(-l) = -4, F'(0) = -2, F'(2) = 2 



lim 2xh + !f- 2h = lim (2x + h - 2) 



h->0 



h->0 



3. Stepl: g(t)= iandg(t + h) 



Step 2: 



i _ i 

. g(t + h)-g(t) _ (t + h)2 t2 



(t + h) 2 

/ l 2 -(l + h) 2 



h h 

h(-2t-h) _ -2t-h 
(t + h) 2 t 2 h (t + h) 2 1 2 



t 2 -(t 2 + 2th + h 2 ) _ -2th - h 2 
(t + h) 2 -t 2 -h (t + h) 2 t 2 h 



Step 3: g'(t) = h lim o ^^ = $ = f ; g'(-l) = 2, g'(2) = - ±, g' ( ^3 | 



2 
3x/3 



4. k(z) = ^ and k(z + h) = ^^> => k'(z) = lim 



a-(z+h) _ i_z > 

, 2(z + h) 2z J 



h->0 



lim 
h-»0 



lim 
h->0 



2(z + h)zh h^To 2(z + h)zh 

^:k'i-l) = -i,k'(l) = -i,k'(v^)=-i 



lim 



lim 



h — > o 2 < z + h ' zh h — > 2 ' z + h ' 



5. Stepl: p(0) = ^/Je and p(6> + h) = ^3(0 + h) 

Step 2- P( fl + h >-P< fl > = v^cfl + h)->/3fl = (x^Tsh-v^) _ (y/3flT3h+y3e) = (3 e + 3h)-3e 

h h h (v/3fl + 3h+x/3fl) h(v/30 + 3h+y3e) 



3h 



^VM+lh+V^fl) V'3ff + 3h+ v /3fl 



Step 3: p'(0) = lim . 3 — ^ = ^ ^- - 

F FV y h->0 x/39 + 3h + y/ie s/36+s/3fJ 2x739 



— -P'(l)= A,p'(3) = i,p'^ 



2V3 



2'*" V3^ 2n /2 



6. r(s) = ^/2^+T and r(s + h) = v /2(sTh)~+T => r'(s) = lim 



72s + 2h+ 1 - x/2s+ 1 



lim 
h-»0 



(v^s+h+l- V2s + l) (x/2s + 2h+l + v^s+l) 



h->0 



lim 



(2s + 2h+l)-(2s+l) 



(V'2s + 2h+l + v /2s+l) h^O h( v /2s + 2h+l + v'2s + l) 



lim 



2h 



lim 



h( v /2l+2h+T+v / 2sTT) h-»0 \72s + 2h+l + v/2s+l x/2s + 1 + 72s + 1 2^25 + 1 



73 i 7T ;r'(0)=l,r'(l)=^,r'(i) = ^ 



t c / \ i 3 j f/ i u\ i/i u\3 > dy i- 2(x + h) 3 -2x 3 ,- 2 (x 3 + 3x 2 h + 3xh 2 + h 3 ) - 2x 3 

7. y = f(x) = 2x° and f(x + h) = 2(x + hr =>■ / = lim — r = lim — r 

dx h->0 h h->0 h 



lim 6x 2 h + 6xh 2 + 2h 3 = Hm 



dx 
h(6x 2 + 6xh + 2h 2 ) 



h-»0 



h->0 



lim (6x 2 + 6xh + 2h 2 ) = 6x 2 
h-»0 



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122 Chapter 3 Differentiation 



dr 



lim 

h->0 



H^ + l -[f + l] 



1 lim [(s + h) 3 + 2]-[s 3 + 2] 



h^O 



lim 
h->0 



s 3 + 3s 2 h + 3sh 2 + h 3 + 2-s 3 -2 _ 1 j im h[3s 2 + 3sh + h 2 ] _ l , ; . vi ; , ,„■ ,,.,., , ,.,_., 3 ,'J 



h->0 



lim (3s 2 + 3sh + h 2 ) = § s 2 



h->0 



9. s = r(t) = ^-r and r(t + h) 



t+h 
2(t+h)+l 



ds 
dt 



lim 
h->0 



(,2(t + h)+lJ (21+1) 



lim 
h->0 

lim 
h->0 



' (t + h)(2t+l)-t(2t + 2h+l) N 
, (2t + 2h+l)(2t+l) 1 



lim 



(t + h)(2t+l)-t(2t + 2h+l) 



1 



2t 2 + 1 + 2ht + h - 2t 2 - 2ht - t 
(2t + 2h+l)(2t+l)h 

_ 1 



q (2t + 2h+l)(2t+l)h 

lim h _ .,.,, 

IfZTd (2t + 2h+l)(2t+l)h h^To (2t + 2h + l)(2t+l) 



lim 



(2t+l)(2t+l) (2t+l) 2 



10. 



dv 



lim 
h->0 



»+•»-!& -('- 



lim 



ht 2 + h 2 t + h 



lim 



— = lim — ^ 
h->0 h 

t 2 + ht + 1 _ t 2 + 1 _ 



lim 
h-»0 



' h(t + h)t-t + (t + hA 

v (t + h)t j 



h" h (' + h > t h" (' + ••)« 



11. p = f(q) = -^i- and f(q + h) 



v^q + hjTT 



dp _ jj m W(q + h) + i7 l^q+T/ 
d 1 h -> h 



lim 
h->0 



>/q + 1 - \/i + '' + 1 

\/q + h+l ^q+1 



lim 



/q+l - y/q + h+l 



h h ^ h^q + h+1 Vq+1 

lim (v/q^-y/q + h + i) . (\/q+7+\/q+h+T) = lim 



(q+l)-(q + h+l) 



h->[) h ^q + h + 1 ^/q + 1 (^q+l + ^/q+h+l) h ^ h^q + h+l ^/q + 1 (^q + 1 + ,/q + h + l) 



lim 



-h 



lim 



h^o h x / q +h+i % /q + i(\/q + 1 + \/q + h + 1 ) h->o ^q+h+i v / q + i( v /q+i + Vq+ h + 1 ) 
^i -1 

yq + 1 yq - 1 ( s/q + 1 - \/q + ' ) 2 (q+ ') \Ai- ' 



12. ^ = lim 

dw 



^3(w + h)-2 ^3w-2J _ j im ^3w - 2 - V3w + 3h - 2 

h^O h h^O h v /3w + 3h-2 v / 3w-2 

(v/3w-2- v/3w + 3h-2~) (' % /3w-2+ v /3w+3h-2~) 

h-»0 h v /3w + 3h-2V3w-2 (^/3w-2 + y/3w + 3h-2\ 

lim (3w-2)-(3w + 3h-2) 

h->0 h v/3w + 3h - 2 ./3w - 2 (y3w - 2 + ./3w + 3h - 2) 

lim 



-3 



I) !) v/3w + 3h - 2 \/3w - 2 (V 3w - 2 + ^/3w + 3h - 2) v/3w - 2 ^/3w - 2 (y3w - 2 + \/3w - 2) 

-3 



2(3w - 2) V3W-2 



13. f(x) = x + 2 and f(x + h) = (x + h) 



f(x + h)-f(x) _ [(" + h )+(7Th)] -[" + !] 



x(x + h) 2 + 9x - x 2 (x + h) - 9(x + h) _ x 3 + 2x 2 h + xh 2 + 9x - x 3 - x 2 h - 9x - 9h _ x 2 h + xh 2 -9h 
x(x + h)h x(x + h)h x(x + h)h 

h(x 2 + xh-9) _ .^±xj^9 ;f , (x)= ^ 4±^_9 = S 2 _9 = 1 _ 9 ;m = f( _ 3) = 



x(x + h)h 



x(x + h) 



h ^0 x < x + h > 



14. k(x) = ^ and k(x + h) = ^^ 



lim 



(2 + x) - (2 + x + h) 



lim 



k'(x)= lim k(x + h >- k(x) 
h->0 h 

lim 



lim 

h->0 



^2+x+h 2+ 



h) 



h" h(2 + x)(2 + x + h) h x ^' h(2 + x)(2 + x + h) h ^"g (2 + x)(2 + x + h) (2 + x)'- 



k'(2) 



16 



15. 



ds 



lim 
h->0 



[(t + h) 3 -(t + h) 2 ]-(t 3 -t 2 ) 



lim 

h->0 



(t 3 + 3t 2 h + 3th 2 + h 3 ) - (t 2 + 2th + h 2 ) - 1 3 + 1 2 



lim 

h->0 



3t 2 h + 3th 2 + h 3 -2th-h 2 



Hm h(3t 2 + 3 t h + h 2 -2 t -h) = nm (3t2 3th h2 _ 2t _ h) 
h^O h h->0 



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Section 3.1 The Derivative of a Function 123 



3t 2 - 2t; m : 



ds I 

dt I i=-l 



16- g 



lim 
h->0 



(x + h+l) 3 -(x+l) 3 



lim 
h->0 



(x + l) 3 + 3(x + l) 2 h + 3(x + l)h 2 + h 3 -(x + l) 3 



lim [3(x + l) 2 + 3(x + l)h + h 2 ] = 3(x + l) 2 ; m 



h-»0 



dy 
dx 



17. f(x) 



and f(x + h) 



V(x + h)-2 



8 (\/x-2- v/x + h-2) (y/x-2 + Vx + h-2) 



f(x + h) - f(x) _ y/|x + h)-2 ^~2 

h ~~ h 



8[(x-2)-(x + h-2)] 



h v /x + h-2 v / x-2 (Vx-2 + Vx + h-2) hv/x + h - 2 V x - 2 ( \/x - 2 + \A + h - 2 ) 



-8h 



h\/x + h-2\/x-2(v'x-2 + \/x + h-2j 

-8 -4 



f'(x)= lim r 8 =■, 

h^O \/x + h - 2 Vx - 2 (yx - 2 + y/x + h - 2 J 

- ; m = f (6) = jtj = — \ =>■ the equation of the tangent 



Vx - 2 ^x - 2 ( Vx - 2 + Vx - 2 j (x-2)Vx 

line at (6, 4) is y - 4 = - ± (x - 6) ^ y = - |x + 3 + 4 ^ y = - ±x + 7 



( 1 + ^4 - (z + h)) - f 1 + \A^z) 

18. g'(z) = lim - — ^ — i = lim 

6 w h->0 h h->0 



\£ 



lim 



(4 - z - h) - (4 - z) 



lim 



4-z-h-\/4-z 



lim 



[V4-z-h+v4-zJ 

(V 4 - z - h + ./4^) 



h^O h(V4-z-h+\/4-z) h^O h (^4- z-h+ ^4- z) h^O (\/4-z - h+ ^4- z) 



m = g'(3) 



2V4-3 



i => the equation of the tangent line at (3, 2) is w — 2 = — \ (z — 3) 



w 



-Iz+f + 2=>w=-iz+|. 



19. s = f(t) = 1 - 3t 2 and f(t + h) = 1 - 3(t + h) 2 = 1 - 3t 2 - 6th - 3h 2 => % = lim 



f(i + h)-f(t) 



h->0 



lim 
h->0 



(l-3t 2 -6th-3h 2 )-(l-3t 2 ) 



lim (-6t - 3h) = -6t 



— I =6 

dt I t=-i ° 



20. y = f(x) = 1 - I and f(x + h) = 1 - ^ => & =-. lim f(x + h , ) ~ f(x) = lim 



dx 
1 L_ 

lim x , x+h = lim , ''.,. = lim -A^x 

h->0 h h->0 "(x + h)h h ^ x(x + h) 



h^O 



h-»0 



dy 
dx 



-Si 



21. r = f(0) 



and f(0 + h) 



lim 



'4-1 
2\/4-8-2\/4-e-h 



x/4-(9 + h) 



dr _. j im f(9 + h)-f(9) _ lim y-4-fl-h ^9 



h->0 



h->0 



_ ini| 2^4^8-2^4^^ (2\/4~^ + 2V4~g-h) 

^""u hv/4^9v/4-9-h h'Tu hy/4^6y/4-B-h ' (2V^fl + 2V4-e-h) 



lim 



lim 



4(4 - 9) - 4(4 - 9 - h) 



lim 



h^O 2h ^4 - 9 ^4 - 8 - h ( ^4 - 6 + y/4 - 8 - h) h^O \/4-0 ^4-8 -h(i/4-8 + \/4-8 -h) 
2 _ 1 . dr I _ 1 

~^ de\g=o 8 



(4-9) (2^4-9) (4~9)V4- 



22. w = f(z) = z + ypL and f(z + h) = (z + h) + y/z + h 

(z + h + v/z + h) - (z + yz) 



dz 



lim 
h->0 



f(z + h) - f(z) 



lim 
1 + lim 



(z + h)-z 



h-»0 h[Jz + h- 



■y/i) 



lim 
h->0 

1 + lim 



h+ Jz + h-Jz 



h->0 Vz + h+^z 



lim 

h->0 



2,/z 



dw I 5 

dz I z=4 4 



/z + h + ,/z 



-h+v/z") 



23. f'(x) = lim 



f(z) ~ f(x) 



lim 



z — > x z-x z^x z-x 



lim 



(x + 2)-(z + 2) 



lim 



lim 



z^»x(z-x)(z + 2)(x + 2) z ^»x(z-x)(z + 2)(x + 2) z " x(z + 2)(x + 2) 



(x + 2)^ 



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124 Chapter 3 Differentiation 



_j 1 

fi2 (\_n2 



24. f (x) = lim fi^I^iW = Hm fr=ff Eg = i im (»-i)Mz-i)' = Hm 

v ' z^x z-x z — ► x z-x z — > x(z-x z-ir x- 1)- z^x 



[{x . 1) . {z . lMx . 1) + {z . 1)] 

2/ ,%2 



lim 



(x — z)(x + z — 2) 



lim 



(z-x)(z-inx-l)- z^x (z-x)(z-l)^(x-l) 

l(x + z-2) _ -l(2x-2) _ -2(x-l) _ -2 



z^x(z-x)(z-l)"(x-l)^ z^x(z-l)"(x-ir (x-1) 4 (x-1) 4 (x-1) 



25. g'(x) = lim 



g(z) - g(x) 



lim 



z(x-l)-x(z-l) 



1 j JQ z-1 »-l _ llln __ |1|N _ |u|1 

z^x z-x z —> x z-x z — > x(z-x)(z- l)(x- 1) z — > x(z — x)(z-l)(x-l) z — > x(z- l)(x- 1) 



lim 



lim 



(x-ir 

26. g'(x) = lim g(z) ' g(x) 



z^x z-x z^x z-x 



lim <i +> A)-(i+^ = lim ^/w^ . a^ = lim — z-x 

' ' z— >x z-x ^z + ^x z — » x(z-x)(,/z + v/x) 



= lim r L t- = ^V 

z — » x s/z + ^/x 2^/x 

27. Note that as x increases, the slope of the tangent line to the curve is first negative, then zero (when x = 0), 
then positive =S> the slope is always increasing which matches (b). 

28. Note that the slope of the tangent line is never negative. For x negative, f^(x) is positive but decreasing as x 
increases. When x = 0, the slope of the tangent line to x is 0. For x > 0, f^(x) is positive and increasing. This 
graph matches (a). 

29. fs(x) is an oscillating function like the cosine. Everywhere that the graph of f3 has a horizontal tangent we 
expect fg to be zero, and (d) matches this condition. 

30. The graph matches with (c). 



31. (a) f is not defined at x = 0, 1, 4. At these points, the left-hand and right-hand derivatives do not agree. 



slope of line joining (—4, 0) and (0, 2) = I but lim 



For example, lim fW ~^ (0) 

F x -> o- x - ° 

line joining (0, 2) and (1, —2) = —4. Since these values are not equal, f'(0) = lim 



f(x) - f(0) 



x^0 



(b) 



> 




> 


. 






3 


_ /'on (-4, 6) 


2 


o-o 


, , <^-A 


'o 1 6 ' ' > 


-8-6-4-2 


_ 2 4 6 8 







r 



x^0 



+ x-0 

f(x) - f(0) 
x-0 



slope of 



does not exist. 



32. (a) 




(b) Shift the graph in (a) down 3 units 

y 




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Section 3.1 The Derivative of a Function 125 



33. 



84 ,«, 87 



-3.3 - 0-<i 



34. (a) 



dp/dt 

20 



x 16 
n 

S 8 




t (days] 



10 20 30 40 50 



(b) The fastest is between the 20 lh and 30" 1 days; 
slowest is between the 40 lh and 50 lh days. 



35. Left-hand derivative: For h < 0, f(0 + h) = f(h) = h (using y = x curve) =4> lim_ 

h — > 



lim 
h->(T 



h 2 -0 



lim h = 0; 

twO" 



Right-hand derivative: For h > 0, f(0 + h) = f(h) = h (using y = x curve) 



lim 
h^O 4 



h-0 



lim 1 = 1; 
h->0+ 



f(0 + h)-f(0) 



f(0 + h) - f(0) 



lim 

lwO+ 



f(0 + h) 


-f(0) 


h 




f(0 + h) - 


f(0) 



Then Inn !lif ' 4i = " la + , lim^ " U + "^ 1W => the derivative f'(0) does not exist. 



h->(r 



h->0+ 



36. Left-hand derivative: When h < 0, 1 + h < 1 => f(l+h) = 2 

= lim = 0; 

h-»rr 



h — (T h 



lim 
h->(T 



Right-hand derivative: When h > 0, 1 + h > 1 => f(l + h) = 2(1 + h) = 2 + 2h => lim 

h — > + 



f(l+h)-f(l) 
h 



lim 



Then lim 
h-»(T 



£±|^ = Um a = lim 2 = 2 

h h^0+ h h->0 + 

f(l+h)-f(l) j_ ,- m f(l+h)-f(l) 



^ lim 
h->0 



the derivative f'(l) does not exist. 



37. Left-hand derivative: When h < 0, 1 + h < 1 => f(l + h) = x/l+h => lim f(1 + h . ) ~ f(1) 

v h -> 0" h 



lim 

IwO" 



'1 +h-l 



lim 
h->(T 



'1+h-lJ 
h 



lim 



(l+h)-l 



lim 



1+h+l) h->(T hfVl+h + l) h->(T v/T+h+l 2 ' 



Right-hand derivative: When h > 0, 1 -I- h > 1 => f(l + h) = 2(1 + h) - 1 = 2h + 1 => lim 



f(l+h)-f(l) 



l im < 2h +')- 1 = lim 2 = 2; 



h->0+ 



h->0+ 



f(l+h)-f(l) _j_ lim f(l + h)-f(l) 
h-0+ h 



Then lim ^=pi^ =£ lim 

h -> o- h 



the derivative f'(l) does not exist. 



38. Left-hand derivative: lim 
h->o _ 

Right-hand derivative: lim 

5 h-0+ 



f(l+h)-f(l) 



f(l+h)-f(l) 



lim 
= lim 

h->0+ 



(l+h)-l 



lim 1 = 1; 

h->0~ 



lim - — j— - 
h^0+ h 



lim 



-h 



-1 



lim 



-1; 



Then lim f(1 + h >- f(1) ± lim f(1 + h >- f(1) 



h-»(T 



h->(T 



the derivative f'(l) does not exist. 



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126 Chapter 3 Differentiation 

39. (a) The function is differentiable on its domain — 3 < x < 2 (it is smooth) 

(b) none 

(c) none 

40. (a) The function is differentiable on its domain — 2 < x < 3 (it is smooth) 

(b) none 

(c) none 

41. (a) The function is differentiable on — 3 < x < and < x < 3 

(b) none 

(c) The function is neither continuous nor differentiable at x = since lim f(x) ^ lim f(x) 

x -> x -» 0+ 

42. (a) f is differentiable on -2 < x < -1, -1 < x < 0, < x < 2, and2 < x < 3 

(b) f is continuous but not differentiable at x = — 1 : lim f(x) = exists but there is a corner at x 

X — » — 1 



1 since 



lim f( - 1+h) - f '- 1) 
h->u" 



-3 and lim 

h->0+ 



f(-l + h)-f(-l) 



h h -► 0+ h 

(c) f is neither continuous nor differentiable at x = and x = 2: 



f'(— 1) does not exist 



at x = 0, lim f(x) = 3 but lim f(x) = = 

x -> 0" x -> 0+ 

at x = 2, lim f(x) exists but lim f(x) d f(2) 

x^2 x->2 w ^ v 



lim f(x) does not exist; 

x — > 



43. (a) f is differentiable on -1 < x < and < x < 2 

(b) f is continuous but not differentiable at x = 0: lim f(x) = exists but there is a cusp at x = 0, so 

x — ► 



f'(0) = lim 



f(0 + h)-f(0) 



h->0 



does not exist 



(c) none 



44. (a) f is differentiable on -3 < x < -2, -2 < x < 2, and 2 < x < 3 

(b) f is continuous but not differentiable at x = —2 and x = 2: there are corners at those points 

(c) none 



45. (a) f'(x) = ,lim f(x + tl ^ f(x) 



(b) 



h->0 

y 

l 



lim 
h->0 



-(x + h) 2 -(-x 2 ) 
h 



lim 

h->0 



-x 2 — 2xh — h 2 + x 2 



lim (— 2x — h) 

h->0 



-2x 




y = -x 



-2x 



-1 



(c) y' = — 2x is positive for x < 0, y' is zero when x = 0, y' is negative when x > 

(d) y = —x 2 is increasing for — oo < x < and decreasing for < x < oo; the function is increasing on intervals 
where y' > and decreasing on intervals where y' < 



46. (a) f'(x) = lim 



f(x + h)-f(x) 



h->0 



lim 



i -i -n 

i X+Jl x ; 



lim 



-x + (x + h) 



h — > x ( x + h ) h 



lim ; , , , = -4 
h — » o x ( x + h ) x 



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Section 3.1 The Derivative of a Function 127 



(b) 



y = ■ 




(c) y' is positive for all x ^ 0, y' is never 0, y' is never negative 

(d) y = — - is increasing for — oo < x < and < x < oo 



47. (a) Using the alternate formula for calculating derivatives: f'(x) = lim — — — = lim -^ - 



_ Hm (z-x)(z 2 + zx + x 2 ) = Um 
z — > x 3(z - x) z — > x 3(z - x) z — ► x 



lim z3 - x3 



f'(x) = X 2 



(b) 





(c) y' is positive for all x/0, and y' = when x = 0; y' is never negative 

(d) y = y is increasing for all x ^ (the graph is horizontal at x = 0) because y is increasing where y' > 0; y is 
never decreasing 



M - jjto 



48. (a) Using the alternate form for calculating derivatives: f'(x) = lirn '""_ " " - - Jim 

z ^x ^ = m (z ' x)(z3 4 r- 2 xr 2z+x3 - -x 7 r 

y 



z^x z-x z^x z-x 

i = lim (*-*)(z 3 +Xz2+A+x3) = JJ Z 3 + XZ3 + A+X3 = X 3 ^ f'(x) = X 3 

z — > x 4(z - x) z — > x 4(z - x) z-tl J 




(c) y' is positive for x > 0, y' is zero for x = 0, y' is negative for x < 

(d) y = x is increasing on < x < oo and decreasing on — oo < x < 



(x— c) (x 2 + xc + c 2 ) 



49. y' = lim ^^^ = lim ^^ = lim "- CMX ' + xc + r) = lim (x 2 + xc + c 2 ) = 3c 2 

J x^c x-c x^c x-c x^c x-c x ^ c v / 

The slope of the curve y = x 3 at x = c is y' = 3c 2 . Notice that 3c 2 > for all c =>• y = x 3 never has a negative 
slope. 



2,/x 4- h — 2 /x 

50. Horizontal tangents occur where y' = 0. Thus, y' = lim — — r — — 
b J J h->0 h 

= lim 2 (^^) (vgh+^) = lim ^((x + hj-x)) = Hm 2 i 

h->0 h (v/xTh+v/x"] h->0 h^Vx + h+v^J h -> \/x + h + \A V x 

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128 Chapter 3 Differentiation 

Then y' = when A= = which is never true => the curve has no horizontal tangents. 



51. y' = lim 
1 h->0 



(2(x + h) 2 - 13(x + h) + 5) - (2x 2 - 13x + 5) 
h 



lim 

h->0 



2x 2 + 4xh + 2h 2 - 13x - 13h + 5 - 2x 2 + 13x - 5 



lim 
h->0 



4xh + 2h 2 -13h 



lim (4x + 2h — 13) = 4x — 13, slope at x. The slope is —1 when 4x — 13 
h->0 



•1 



> 4x = 12 => x = 3 => y = 2-3 2 — 13-3 + 5= 16. Thus the tangent line is y + 16 = (-l)(x - 3) 
y = — x — 13 and the point of tangency is (3, —16). 



52. For the curve y = -Jx, we have y' 



lim 
h->0 



(Vx + h- x/x) (\A + h +V / *) 



lim 



(x + h) - x 



(\A + h + yit) h^O (\/x + h+x/x) h 



lim , } — y= - 

h^O V x + h +V x 2^/x 



y- . Suppose (a, ^/a) is the point of tangency of such a line and (— 1, 0) is the point 



on the line where it crosses the x-axis. Then the slope of the line is ^zrz\\ = irj which must also equal 



. - tV => 2a = a + 1 => a=l. Thus such a line does 

a+ 1 2\/a 



t^ ; using the derivative formula at x = a 

exist: its point of tangency is (1, 1), its slope is — k= = ^; and an equation of the line is y — 1 = | (x — 1) 

^y=ix+i. 

53. No. Derivatives of functions have the intermediate value property. The function f(x) = |_xj satisfies f(0) = 
and f(l) = 1 but does not take on the value | anywhere in [0, 1] =>■ f does not have the intermediate value 
property. Thus f cannot be the derivative of any function on [0, 1] => f cannot be the derivative of any function 
on (—oo, oo). 



54. The graphs are the same. So we know that 
for f(x) = |x| , we have f'(x) = — . 



1 I 



-i 



<H 



55. Yes; the derivative of — f is — f so that f'(xo) exists =>• — f'(xo) exists as well. 

56. Yes; the derivative of 3g is 3g' so that g'(7) exists => 3g'(7) exists as well. 



57. Yes, lim fi2 can exist but it need not equal zero. For example, let g(t) = mt and h(t) = t. Then g(0) = h(0) 



0, but lim 



g(0 



lim 



o w t->0 



lim m = m, which need not be zero. 
t-»0 



58. (a) Suppose |f(x)[ < x 2 for -1 < x < 1. Then |f(0)| < 2 => f(0) = 0. Then f'(0) = lim 



gO + h) - f(0) 



h^O 



lim 
h->0 



f(h)-0 



lim 
h->0 



f(h) 



For |h| < 1, -h 2 < f(h) < h 2 =>■ -h < ^ < h => f'(0) = lim 



f(h) 



h->0 







by the Sandwich Theorem for limits, 
(b) Note that for x ^ 0, |f(x)| = |x 2 sin A| = |x 2 | |sinx| < |x 2 | • 1 = x 2 (since -1 < sin x < 1). By part (a), 
f is differentiable at x = and f'(0) = 0. 



59. The graphs are shown below for h = 1, 0.5, 0.1. The function y = =K- is the derivative of the function 

2y x 



/x so that — 7- = lim 

V x h->0 



The graphs reveal that y 



— — r — — gets closer to y = — K 



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Section 3.1 The Derivative of a Function 129 



as h gets smaller and smaller. 

y 




h = 0.5 




h = 0.1 




60. The graphs are shown below for h = 2, 1, 0.5. The function y = 3x 2 is the derivative of the function y = x 3 so 



that 3x 2 = lim (x+h ^ x . The graphs reveal that y 
h-»0 h 

gets smaller and smaller. 



(x+h) 3 -x 3 



gets closer to y = 3x 2 as h 



61. Weierstrass's nowhere differentiable continuous function. 

y 




*M = cosfrx) + Qj cos(9jrjr) + (J\ cosi^nx) + f|Ycos(9V*) 
\j\ cos&irx) 



H 

h-O.ZU 


2 




// 


(x + h)< - X> 

k 


\ 1 


4> 





-10 12 



+ ••• + 



62-61. Example CAS commands: 
Maple : 

f := x -> x A 3 + x A 2 - x; 



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130 Chapter 3 Differentiation 

x0:= 1; 

plot( f(x), x=xO-5..xO+2, coloi-black, 

title="Section 3_1, #62(a)" ); 
q := unapply( (f(x+h)-f(x))/h, (x,h) ); # (b) 

L := limit( q(x,h), h=0 ); # (c) 

m := eval( L, x=x0 ); 
tan_line := f(x0) + m*(x-x0); 
plot( [f(x),tan_line], x=x0-2..x0+3, color=black, 

linestyle=[l,7], title="Section 3.1 #62(d)", 

legend=["y=f(x)", "Tangent line at x=l"] ); 
Xvals := sort( [ xO+2 A (-k) $ k=0..5, xO-2 A (-k) $ k=0..5 ] ): # (e) 

Yvals := map( f, Xvals ): 

evalf[4](< convert(Xvals,Matrix) , convert(Yvals,Matrix) >); 
plot( L, x=x0-5..x0+3, color=black, title="Section 3.1 #62(f)" ); 
Mathematica : (functions and xO may vary) (see section 2.5 re. RealOnly ): 
«Miscellaneous"RealOnly" 
Clear[f, m, x, y, h] 
x0= 7T /4; 
f[x_]:=x 2 Cos[x] 
Plot[f[x], {x,x0 - 3,x0 + 3}] 
q[x_,h_]:=(f[x + h]-f[x])/h 
m[x_]:=Limit[q[x,h], h ->• 0] 
ytan:=f[x0] + m[x0] (x - xO) 
Plot[{f[x], ytan},{x, xO - 3, xO + 3}] 
m[x0 - 1]//N 
m[x0 + 1]//N 
Plot[{f[x], m[x]},{x, xO - 3, xO + 3}] 

3.2 DIFFERENTIATION RULES 



1. y = -x 2 + 3 => 



dy - A f_ x 2^ i A 

dx dx V A ) ~ dx 



(-x 2 ) + £ (3) = -2x + = -2x =>• 



d 2 y 



2. y = x 2 + x + 8 => a = 2x+l+0 = 2x+l =>• 



d 2 v 
dx- 1 



3. s = 5t 3 - 3t 5 =► | = | (5t 3 ) - g (3t 5 ) = 15t 2 - 15t 4 =* § = | (15t 2 ) - | (15t 4 ) = 30t - 60t 3 

4. w = 3z 7 - 7z 3 + 21z 2 =>■ g = 21z° - 21z 2 + 42z =>■ ff = 126z 5 - 42z + 42 



4 v 3 _ v . .. dy _ 4v 2 _ , ^ dfy = gx 



5. y = | x 3 - x ^ ^y = 4x 2 - 1 ^ 



6. y=^ + ^ + | =>. d| =x 2 +x+ l ^ g=2x+l+0 = 2x+l 



7. w = 3z~ 2 -z" 1 => 



1 _, dw 



-6z" 



-6 | 1 



^ = 18z- 4 -2z- 3 =if-| 



8. s = -2T 1 + 4t- 2 => | = 2t- 2 - 8r 3 = | - | =* |§ = -4t- 3 + 24r 4 = ^ + £ 

9. y = 6x 2 - lOx - 5x~ 2 => jjz = I2x - 10 + 10x~ 3 = 12x - 10 + ^ => f$ = 12 - - 30x~ 4 = 12 - 2° 



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Section 3.2 Differentiation Rules 131 



10. y = 4 - 2x - x- 3 => ^ 



-2 + 3x~ 4 = -2+3^g=0- 12x- 5 = -F 



11. r= I s -2_I s -l 



dr 
ds 



2 E -3 i 5 „-2 _ ^2 | _5_ 

3 s t 2 5 ~~ 3s 3 "^ 2s 2 



d^r 
d>- 



2s~ 4 - 5s- 3 = 4 - t 



12. r = 126I- 1 -46>- 3 +6>- 4 

24 48 | 20 



| = -l26>- 2 + 126> 4 46» 



-*> - rJ2 _i_ 12 _ £ 

" 92 T 04 05 



|§ = 246»- 3 - 48<r 5 + 206>" 6 



13. (a) y = (3 - x 2 ) (x 3 - x + 1) => y' = (3 - x 2 ) • £ (x 3 - x + 1) + (x 3 - x + 1) - £ (3 - x 2 ) 

= (3 - x 2 ) (3x 2 - 1) + (x 3 - x + 1) (-2x) = -5x 4 + 12x 2 - 2x - 3 
(b) y = -x 5 + 4x 3 - x 2 - 3x + 3 => y 1 = -5x 4 + 12x 2 - 2x - 3 

14. (a) y = (x - 1) (x 2 + x + 1) => y' = (x - l)(2x + 1) + (x 2 + x + 1) (1) = 3x 2 
(b) y = (x - 1) (x 2 + x + 1) = x 3 - 1 => y' = 3x 2 

15. (a) y = (x 2 + l)(x + 5+i) => y' = (x 2 + l)-£(x + 5+i) + (x + 5 + i)-£(x 2 + l) 

= (x 2 + 1) (1 - x- 2 ) + (x + 5 + x- 1 ) (2x) = (x 2 - 1 + 1 - x- 2 ) + (2x 2 + lOx + 2) = 3x 2 + lOx + 2 - X 2 
(b) y = x 3 + 5x 2 + 2x + 5 + i =>■ y' = 3x 2 + lOx + 2 - 4j 



16. y=(x+I)(x-I + l) 

(a) y' = (x + x- 1 ) - (1 + x- 2 ) + (x - x" 1 + 1) (1 - x~ 2 ) = 2x + 1 - A + | 



(b) y = x 2 + x + 1 - A. => y' = 2x + 1 - i + 



x 2 + x 3 



17. y = f^l ; use the quotient rule: u = 2x + 5 and v = 3x - 2 =>■ u' = 2 and v' = 3 =$■ y' = V "'~ 2 UV ' 

_ (3x-2)(2)-(2x + 5)(3) _ 6x-4-6x-15 _ -19 



(3x - 2) 2 



(3x-2) 2 (3x-2) 2 



18. z 



2x+l . dz _ (x 2 -l)(2)-(2x+l)(2x) _ 2x 2 - 2 - 4x 2 - 2x _ -2x 2 - 2x - 2 _ -2(x 2 + x+l) 



x 2 — 1 dx 



(x 2 -l) 2 



(x 2 -l) 2 



(x 2 -l) 2 



(" 2 -ir 



19. g(x) = |tq| ; use the quotient rule: u = x 2 — 4 and v = x + 0.5 => u' = 2x and v' = 1 =>■ g'(x) 

_ (x + 0.5)(2x)-(x 2 -4)(l) _ 2x 2 + x-x 2 + 4 _ x 2 + x + 4 



(x + 0.5) 2 



(x + 0.5) 2 (x + 0.5) 2 



t 2 -l _ (t-l)(t + l) _ t+1 
t 2 + t-2 ~~ (t- 



- U - 1!! » - ,. i i- > - Ct+2)(t-l) ~~ t+2' L V^ 1 =* I (V) 



t(t\ - (t + 2)(l)-(t+l)(l) = t + 2-t-l = 1 

(t + 2) 2 (t + 2) 2 (t + 2) 2 



21. V = (l-t)(l+t 2 r 1 = i^ => dx = (l+t 2 )(-l)-0-'>(2t) _ -l- t 2 -2t + 2t 2 _ t^jt^l 



22. w 



x + 5 
2x-7 



(1+t 2 ) 2 ~ (1+t 2 ) 2 (1+t 2 ) 2 

I _ (2x-7)(l)-(x + 5)(2) _ 2x - 7 - 2x - 10 _ -17 



(2x - 7) 2 



(2x-7) 2 (2x-7) 2 



23. f(s) 



vA-i 



f'(s) 



(y^ 1 )^)-^-')^) _ (^+i)-(^-i) 



v^+i " ' '" ~ (Vi+i) 2 2^(^+1)" 



^(%A+i) 2 



NOTE: ^ (,/s) = ^V from Example 2 in Section 2.1 



24. u 



5x+l 
2~7x 



du 
dx 



(2 v ^)(5)-(5x+l)(-ij) 



4x 



5x- 1 

4x 3 / 2 



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132 Chapter 3 Differentiation 



25. v 



1 +x-4Jx 



£ x(l-^)-(l+x-V3 _ 2 ^_ 



26. r = 2 



ITS 



of v^o)-i(^) ! 



1 

a 12 



27. y = ( x 2_ 1 wx 2 + x+n > use tne quotient rule: u = 1 and v = (x 2 — 1) (x 2 + x + 1) => u' = and 

v' = (x 2 - 1) (2x + 1) + (x 2 + x + 1) (2x) = 2x 3 + x 2 - 2x - 1 + 2x 3 + 2x 2 + 2x = 4x 3 + 3x 2 - 1 



dy vu' — uv' 

dx v 2 



0-l(4x 3 + 3x 2 -l) 



-4x 3 - 3x 2 + 1 



(x 2 - l) 2 (x 2 + x+ 1)" (x 2 - If (x 2 + x+ 1)" 



28. y 



(x+l)(x + 2) x 2 + 3x + 2 . / (x 2 - 3x + 2) (2x + 3) - (x 2 + 3x + 2) (2x - 3) _ -6x 2 + 12 
(x-l)(x-2) x 2 -3x + 2 = ^ J (x-l) 2 (x-2) 2 (x-l) 2 (x-2) 2 

-6 (x 2 - 2) 
(x-l) 2 (x-2) 2 



29. y 

30. y 

31. y 

32. S: 



1 „4 3 v 2 



x 2 - x => y' = 2x 3 - 3x - 1 => y" = 6x 2 - 3 => y'" = 12x =4> y (4) = 12 =>• yM = for all n > 5 



i-x 5 =^ y' 



120 



x J +7 



24 A 



v " _ I v 3 _^ ..III _ 1 2 
y — , a => y — „ a 



/(4) 



=^ y( 5 ' = 1 => yM = for all n > 6 



x 2 + 7X- 1 =► p- = 2x - 7x- 2 = 2x - \ 

dx x 2 



g=2+14x- 3 = 2+i| 



^±|M = 1 + f - P = 1 + St" 1 - r 2 =► |=0- 5r 2 + 2r 3 = -5r 2 + 2r 3 = ^ + | 

d 2 ! _ -, n t -3 _ fit -4 _ 10 _ 6 

dt 2 _ lul Dt - 7 P 



33. r: 



(g-D(fl 2 + g+i) 

3 



W 3 



3-3 



^ = + 3(T 4 = 3<T 4 



A , eft 

e» ^ do 2 



-i2r 5 = ^ 



34. u 



(x 2 + x)(x 2 -x+l) _ x(x+l)(x 2 -x+l) _ x(x 3 + l) _ x " + > 



1 + A = 1 + x 



-3 



1=0- 3x- 4 



-3x- 4 = =| => & = 12x- 5 = if 

x 4 dx 2 x J 



35. w 



;i±^) (3 - z) = (i z- 1 + 1) (3 - z) = z- 1 - i + 3 - z = z- 1 



dw 



0-1 



d-\v 
dz-' 



2z- 3 - = 2z 



-3 _ 2 



36. w = (z + l)(z - 1) (z 2 + 1) = (z 2 - 1) (z 2 + 1) = z 4 - 1 => f- = 4z 3 - = 4z 3 => 



d 2 w 
dz 2 



12z 2 



37. p 



38. p 



I q 2 + 3 \ / q 4 - 1 \ _ q 6 - q 2 + 3q 4 

V i2q / V i 3 / 12 i 4 

d 2 P _ 1 1 „-4 



-3 1 „2 1 „-2 ill „-4 . dp _ 1 , 1 3 , 5 _ 1 



q '-si 



dq 6 



q + i q + q" 



6q 3 



dq 2 — 6 2 1 

q 2 + 3 



5q 



-6 _ 1 1 

' 6 2q 4 



q 2 + 3 



q 2 + 3 _ q 2 + 3 



J_ - I a -l 

(q-l) 3 + (q+l) 3 ~~ (q 3 -3q 2 + 3q-l) + (q 3 + 3q 2 + 3q+l) — 2q 3 + 6q — 2q (q 2 + 3) — 2q — 2 4 



dp 
dq 



iq- 2 



.J_ =s. ^E _ -3 _ J_ 
2q 2 ^ dq 2 — 4 — q 3 



39. u(0) = 5, u'(0) = -3, v(0) = -1, v'(0) = 2 

(a) £ (uv) = uv' + vu' =* £ (uv)| x=o = u(0)v'(0) + v(0)u'(0) = 5 - 2 + (-l)(-3) = 13 

d (a\ _ vu'-uv' . d (u\\ _ v(0)u'(0) - u(0)v / (0) _ (-l)(-3) - (5)(2) _ n 



(b) 

(C) 



dx \vJ v 2 ~^ dx VvVlx-o (v(0)) 2 (-1) 2 

_d_ (v\ _ uv' - vu' . _d_ fv\| _ u(0)v'(0) - v(0V(0) _ (5)(2)-(-l)(-3) _ _7_ 

dx I u / — u 2 ^ dx V u ) I x =o ~~ (u(0)) 2 ~~ (5) 2 ~~ 25 



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(d) £ (7v - 2u) = 7v' - 2u' =>• £(7v-2u)| 



Section 3.2 Differentiation Rules 133 

7v'(0) - 2u'(0) = 7 • 2 - 2(-3) = 20 



40. u(l) = 2, u'(l) 



(a) 
(b) 

(c) 
(d) 



s(«v)| 



dx 

dx \uJ 



0, v(l) = 5, v'(l) = - 
u(l)v'(l) + v(l)u'(l) 

v(l)u'(l)-u(l)v'(l) 



2-(-l) + 5-0 



(V(l))2 

u(l)v'(l)-v(l)u'(l) 

(U(l))2 



5-0— 2-(— 1 ) 

(5)2 

2-(-l)-5-0 

(2)2 



2_ 

25 



dx 



(7v - 2u)| _ = 7v'(l) - 2u'(l) = 7 - (-1) - 2 - = -7 



41. 



y = x 3 - 4x + 1. Note that (2, 1) is on the curve: 1 = 2 3 - 4(2) + 1 

(a) Slope of the tangent at (x, y) is y' = 3x 2 - 4 => slope of the tangent at (2, 1) is y'(2) = 3(2) 2 -4 = 8. Thus 
the slope of the line perpendicular to the tangent at (2, 1) is — | =>• the equation of the line perpendicular to 
to the tangent line at (2, 1) is y — 1 = — | (x — 2) or y = — f + f • 

(b) The slope of the curve at x is m = 3x 2 — 4 and the smallest value for m is —4 when x = and y = 1. 



(c) We want the slope of the curve to be 8 => y' = 8 => 3x 2 — 4 = 8 = 
x = 2, y = 1 and the tangent line has equation y — 1 = 8(x — 2) or y 
y = (— 2) 3 — 4(— 2) +1 = 1, and the tangent line has equation y — 1 : 



3x 2 = 12 =^> x 2 = 
8x — 15; when x = 
8(x + 2) or y = 8x 



4 =>• x 

-2, 

1-17. 



±2. When 



42. (a) y 



x 3 — 3x — 2 =4> y' = 3x 2 — 3. For the tangent to be horizontal, we need m = y' = => = 3x 2 — 3 
=>• 3x 2 = 3 =4> x = ±1. When x = — 1, y = => the tangent line has equation y = 0. The line 
perpendicular to this line at (— 1, 0) is x = — 1. When x = 1, y = — 4 =>• the tangent line has equation 
y = —4. The line perpendicular to this line at (1, —4) is x = 1. 
(b) The smallest value of y' is —3, and this occurs when x = and y = —2. The tangent to the curve at (0, —2) 



has slope —3 =>■ the line perpendicular to the tangent at (0, —2) has slope i 

y 



y- 



2 = i (x - 0) or 



43. y 



4x 



It. 
3 * 


- 2 is an equation of the 


perpendicular 


line. 




=>• 


dy 
dx 


_ (x 2 +l)(4)-(4x)(2x) 

(x2+l) 2 


4x 2 + 4-8x 2 

(x2+l)2 


. 4(- 
(x 


x 2 + l 
+ 1) 2 



X 2 +l 

= 4, so the tangent to the curve at (0, 0) is the line y = 4x. When x 
curve at (1, 2) is the line y = 2. 



4(0+1) 
1 



When x = 0, y = and y' 

1 , y = 2 => y' = 0, so the tangent to the 



44. 



(x 2 + 4)(0)-8(2x) 



-16x 



When x = 2, y = 1 and y' 



) x 2 + 4 -r J ~ (x 2 + 4) 2 (x 2 + 4) 2 

line to the curve at (2, 1) has the equation y — 1 = — I (x — 2), or y 



-16(2) 
(2 2 + 4) 2 

1-2. 



5 , so the tangent 



45. y = ax 2 + bx + c passes through (0, 0) =>■ = a(0) + b(0) + c 



2 = a + b; y' = 2ax + b and since the curve is tangent to y = x at the origin, its slope is 1 at x = 
y' = 1 when x = => 1 = 2a(0) + b =>• b = 1. Then a + b = 2 =>- a=l. In summary a = b - 



0; y = ax 2 + bx passes through (1, 2) 

1 and c = so 



the curve is y 



v-2 



46. 



passes through (1, 0) => = c(l) — 1 



1 — 2x and x = 1 



— 1. Since y = x — x 2 and y 



1 =>• the curve is y = x — x . For this curve, 
x 2 + ax + b have common tangents at x = 0, 



x 2 + ax + b must also have slope — 1 at x = 1 . Thus y' = 2x + a =>• — 1 = 2-1 
y = x 2 — 3x + b. Since this last curve passes through (1, 0), we have = 1 — 34 
—3, b = 2 and c = 1 so the curves are y = x 2 — 3x + 2 and y = x — x 2 . 



a => a = — 3 
3 =>• b = 2. In summary, 



47. (a) y 

y 



x° - x =>• y 
2(x + 1) or y 



--3x 2 - 1. Whenx = -l,y 

2x + 2. 



and y' = 2 =>• the tangent line to the curve at (—1, 0) is 



Copyright (c) 2006 Pearson Education 




134 Chapter 3 Differentiation 

(b) 




(c) 



* * I => x 3 - x = 2x + 2 =>■ x 3 - 3x - 2 = (x - 2)(x + l) 2 = 



=>• x = 2 or x = - 1. Since 



y = 2(2) + 2 = 6; the other intersection point is (2, 6) 



48. (a) y = x 3 - 6x 2 + 5x => y' = 3x 2 - 12x + 5. When x = 0, y = and y' = 5 
(0, 0) is y = 5x. 
(b) 



the tangent line to the curve at 



(c) 




y = x 3 — 6x 2 + 5x 



y = 5x 
Since y = 5(6) = 30, the other intersection point is (6, 30). 



6x 2 + 5x = 5x =^ x 3 - 6x 2 = =^ x 2 (x - 6) = =4> x = 0orx = 6. 



49. P(x) = a n x n + a n _!X n 



a 2 x 2 + ajx + a => P'(x) = na n x n : + (n — l)a n _ 1 x l 



n-2 



+ 2a 2 x + ai 



50. R = M 2 (f - f ) = f M 2 - | M 3 , where C is a constant 



dR 



CM-M 2 



51. Let c be a constant =>■ gf = => |^(u-c) = u-^ + c-^=u-0 + c^ = c^. Thus when one of the 
functions is a constant, the Product Rule is just the Constant Multiple Rule =>■ the Constant Multiple Rule is 
a special case of the Product Rule. 

n I dv ■< dv 

52. (a) We use the Quotient rule to derive the Reciprocal Rule (with u=l): f- (-) = T JkL = — 2^ 



v 2 dx 



»• A. (^ 



(b) Now, using the Reciprocal Rule and the Product Rule, we'll derive the Quotient Rule. K v N ; d\ 

= u • E (v) + v • I (Product Rule) = u-(^)^ + ^ (Reciprocal Rule) 



*-f u .V 



dx V v / v 2 



= dx y2 dx , the Quotient Rule. 
53. (a) A (uvw)= A ((uv) . w) = (uv) dw +w d (uv) = uv dw +w(u d^ +v ^ =uv dw +wu d^ +wv d^ 



dx 



uvw' + uv'w + u'vw 



(b) £ (U1U2U3U4) = £ ((U1U2U3) U 4 ) = (U!U 2 U 3 ) |j± + U 4 £ (U1U2U3) 



U lU2 U 3 £ + U 4 ( Ul U 2 £ + U 3Ul £ + U 3 U 2 £; 



dxV— »/ - S (U 1U2 U 3 U 4 ) 

(using (a) above) 



=* A (u lU2 u 3 u 4 ) = u lU2 u 3 to + Ul u 2 u 4 to + Ul u 3 u 4 to + u 2 u 3 u 4 to 
= UiU 2 u 3 u 4 + UiU 2 u 3 u 4 + uiu 2 u 3 u 4 + uiu 2 u 3 u 4 
(c) Generalizing (a) and (b) above, 4- (ur • -u n ) = Uiu 2 - • -u^iu^ + Uiu 2 - • ■u„_ 2 u' 1 _ 1 u 11 + ... + u'^- ■ -u n 



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Section 3.3 The Derivative as a Rate of Change 135 



54. In this problem we don't know the Power Rule works with fractional powers so we can't use it. Remember 
d (\/*) = TT~ (f rom Example 2 in Section 2.1) 



dx 



(a) aH* 3/2 ) = aH* • * 1/2 ) = * • £ (\A) + 



(b) 



^s« = x -2t; 



'x-l 



+ 



/^ = !^ = 3 x l/2 



( X 5/2) = A ( x 2 . x l/2) = x 2 ^ ^ + ^ _d_ (x 2) = x 2 . ^_±_j + ^ ■ 2x = ± X 3 / 2 + 2x 3 / 2 = § X 3 / 2 

( c ) £ ( x7/2 ) = £ ( x3 • xV2 ) = x3 £ (v^) + V 7 ^ £ ( x3 ) = x3 • (jfe) + \A • 3x2 = 3 x5/2 + 3x5/2 = \ x5/2 

(d) We have £ (x 3 / 2 ) = | x 1 / 2 , £ (x 5 / 2 ) = f x 3 / 2 , £ (x 7 / 2 ) = | x 5 / 2 so it appears that £ (x 11 / 2 ) = § x^ 2 )" 1 
whenever n is an odd positive integer > 3. 



55. p 



nRT 



dP 
dV 



V- 



We are holding T constant, and a, b, n, R are also constant so their derivatives are zero 

-nRT 



(V-nb)-0-(nRT)(l) _ V 2 (0) - (an 2 ) (2V) 

(V-nb) 2 (y2) 2 ~~ (V-nb) 2 



2iin- 
V 3 



56. A(q) = s + cm 



hq 
2 



(km)q- 1 +cm+ (|)q=> 



dA 
dq 



-(km)q- 2 +(|) 



km I h 
' q 2 + 2 



dt 2 



2(km)q 



-3 _ 2km 

- q3 



3.3 THE DERIVATIVE AS A RATE OF CHANGE 



t 2 - 3t + 2, < t < 2 



As 
At 



(a) displacement = As = s(2) — s(0) = Om — 2m = —2 m, v !iv - v — 

(b) v = | = 2t - 3 => |v(0)| = |-3| = 3 m/sec and |v(2)| = 1 m/sec; 
a = ff = 2 => a(0) = 2 m/sec 2 and a(2) = 2 m/sec 2 



-1 m/sec 



(c) v = 0=>2t — 3 = => t = |. v is negative in the interval < t < | and v is positive when | < t < 2 =>■ the body 



changes direction at t 



s = 6t - t 2 , < t < 6 

(a) displacement = As = s(6) — s(0) = m, v. lv = 4| = | = m/sec 

(b) v = | = 6 - 2t => |v(0)| = | 6| = 6 m/sec and |v(6)| = |-6| = 6 m/sec; 
a = |f = -2 => a(0) = -2 m/sec 2 and a(6) = -2 m/sec 2 

(c) v = => 6 — 2t = =4> t = 3. v is positive in the interval < t < 3 and v is negative when 3 < t < 6 => the body 
changes direction at t = 3. 



d£s 
dt 2 



-6t + 6 



s = -t 3 + 3t 2 - 3t, < t < 3 

(a) displacement = As = s(3) — s(0) = —9 m, v av = ^| = ^ = — 3 m/sec 

(b) v = | = -3t 2 + 6t - 3 => |v(0)| = |-3| = 3 m/sec and |v(3)| = |-12| = 12 m/sec; a 

=> a(0) = 6 m/sec 2 and a(3) = — 12 m/sec 2 

(c) v = => -3t 2 + 6t - 3 = => t 2 - 2t + 1 = =^> (t - l) 2 = => t = 1. For all other values of t in the 
interval the velocity v is negative (the graph of v = — 3t 2 + 6t — 3 is a parabola with vertex at t = 1 which 
opens downward => the body never changes direction). 



4. s 



f 3 i a 



t 2 , < t < 3 



(a) As = s(3) - s(0) 



m, v.,. 



As 
At 



m/sec 



(b) v = t 3 - 3t 2 + 2t => |v(0)| = m/sec and |v(3)| = 6 m/sec; a = 3t 2 - 6t + 2 => a(0) = 2 m/sec 2 and 
a(3) = 11 m/sec 2 

(c) v = => t 3 - 3t 2 + 2t = => t(t - 2)(t - 1) = => t = 0, 1, 2 => v = t(t - 2)(t - 1) is positive in the 
interval for < t < 1 and v is negative for 1 < t < 2 and v is positive for 2 < t < 3 => the body changes direction at 



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136 Chapter 3 Differentiation 

t = 1 and at t = 2. 



5. s= f - 5 1 < t<5 

v- t ' — — 



(a) As = s(5) - s(l) = -20 m, v a 



-20 
4 



-5 m/sec 



150 10 

t 4 ,3 



a(l) = 140 m/sec 2 and 



(b) v = ^|° + | => |v(l)| = 45 m/sec and |v(5)| = \ m/sec; a 
a(5) = j~ m/sec 2 

(c) v = =4> ~ 5 ° 3 +5t = =4> -50 + 5t = =>• t = 10 => the body does not change direction in the interval 



6. s= ^, -4 < t < 

(a) As = s(0) - s(-4) = -20 m, v av 



20 



-5 m/sec 



(b) v 



-25 
(t+5) 2 



l v (~ 4)| = 25 m/sec and |v(0)| = 1 m/sec; a 



50 

(t+5) 3 



a(— 4) = 50 m/sec 2 and 



a(0) = % m/sec 2 



(c) v = => 25 )2 =0 =>- v is never => the body never changes direction 

s = t 3 — 6t 2 + 9t and let the positive direction be to the right on the s-axis. 

(a) v = 3t 2 - 12t + 9 so that v = =>■ t 2 -4t + 3 = (t- 3)(t - 1) = =>• t = 1 or 3; a = 6t - 12 => a(l) 

= —6 m/sec 2 and a(3) = 6 m/sec 2 . Thus the body is motionless but being accelerated left when t = 1, and 
motionless but being accelerated right when t = 3. 

(b) a = =>■ 6t - 12 = => t = 2 with speed | v(2)| = 1 12 - 24 + 9| = 3 m/sec 

(c) The body moves to the right or forward on < t < 1, and to the left or backward on 1 < t < 2. The 
positions are s(0) = 0, s(l) = 4 and s(2) = 2 =4> total distance = |s(l) - s(0)| + |s(2) - s(l)| = |4| + |-2| 

= 6 m. 

v = t 2 -4t + 3 =4> a = 2t-4 

(a) v = => t 2 - 4t + 3 = =>• t = 1 or 3 => a(l) = -2 m/sec 2 and a(3) = 2 m/sec 2 

(b) v > =>■ (t - 3)(t - 1) > =>■ < t < 1 or t > 3 and the body is moving forward; v < =*> (t - 3)(t - 1) < 

=> 1 < t < 3 and the body is moving backward 

(c) velocity increasing =4> a>0 => 2t-4>0 => t>2; velocity decreasing =4> a<0 =>• 2t-4<0 => < t < 2 



1.86t 2 



3.72t and solving 3.72t = 27.8 => t « 7.5 sec on Mars; s i = 11.44t 2 



22.88t and 



solving 22.88t = 27.8 =^ t w 1.2 sec on Jupiter. 



10. (a) v(t) = s'(t) = 24 - 1.6t m/sec, and a(t) = v'(t) = s"(t) = -1.6 m/sec 2 

(b) Solve v(t) = => 24 - 1.6t = => t = 15 sec 

(c) s(15) = 24(15) -. 8(15) 2 = 180m 

(d) Solve s(t) = 90 => 24t - .8t 2 =90 => t = 30± ' 2 5v ^ w 4.39 sec going up and 25.6 sec going down 

(e) Twice the time it took to reach its highest point or 30 sec 

11. s = 15t - \ g s t 2 => v = 15 - g s t so that v = => 15 - g s t = => g s = l -f ■ Therefore g s = ^ = | = 0.75 m/sec 2 

12. Solving s m = 832t - 2.6t 2 = => t(832 - 2.6t) = => t = or 320 => 320 sec on the moon; solving 

s e = 832t - 16t 2 = =^ t(832 - 16t) = => t = or 52 =^> 52 sec on the earth. Also, v m = 832 - 5.2t = 
=> t= 160 and s m (160) = 66,560 ft, the height it reaches above the moon's surface; v e = 832 — 32t = 
=>• t = 26 and s e (26) = 10,816 ft, the height it reaches above the earth's surface. 



13. (a) s = 179 - 16t 2 



-32t => speed = |v| = 32t ft/sec and a 



-32 ft/sec 2 



Copyright (c) 2006 Pearson Education 




Section 3.3 The Derivative as a Rate of Change 137 



(b) s = =$> 179 - 16t 2 = =>■ t 



(c) Whent 



179 



-32 



179 



"if « 3.3 sec 
-8\/l79« -107.0 ft/sec 



14. (a) lim v = lim 9.8(sin 9)t = 9.8t so we expect v = 9.8t m/sec in free fall 



"2 "2 

(b) a = f t = 9.8 m/sec 2 



15. (a) at 2 and 7 seconds 
(c) 



\v\ (m/sec) 



(b) between 3 and 6 seconds: 3 < t < 6 

(d) 



Speed 



■ ' ' V i i i i V i i \ , , (sec) 



2 4 6 8 10 



4 - 


dv 


3 


— o 


<^-^ 


2 








, 







_ 1 


23456789 10 


-2 




o^— 


-3 


-<^—o 


-4 







16. (a) P is moving to the left when 2<t<3or5<t<6;Pis moving to the right when < t < 1; P is standing 
still when l<t<2or3<t<5 

(b) 



(cm/sec) 



i r 

2 



velocity 



-o— O — O o t (sec) 



12 3 5 6 



speed (cm/sec) 

4 



-O— o — o o 

12 3 5 6 



t (sec) 



17. (a) 190 ft/sec 

(c) at 8 sec, ft/sec 

(e) From t = 8 until t = 10.8 sec, a total of 2.8 sec 

(f) Greatest acceleration happens 2 sec after launch 

(g) From t = 2 to t = 10.8 sec; during this period, a = v( '° 8 8 ) :^ (2) 



(b) 2 sec 

(d) 10.8 sec, 90 ft/sec 



-32 ft/sec 2 



18. (a) Forward: < t < 1 and 5 < t < 7; Backward: 1 < t < 5; Speeds up: 1 < t < 2 and 5 < t < 6; 
Slows down: < t < 1, 3 < t < 5, and 6 < t < 7 

(b) Positive: 3 < t < 6; negative: < t < 2 and 6 < t < 7; zero: 2<t<3and7<t<9 

(c) t = and 2 < t < 3 

(d) 7 < t < 9 



19. s = 490t 2 



v = 980t => a = 980 



s(4/7)-s(0) 
4/7 



280 cm/sec. 



(a) Solving 160 = 490t 2 => t = I sec. The average velocity was 

(b) At the 160 cm mark the balls are falling at v(4/7) = 560 cm/sec. The acceleration at the 160 cm mark 
was 980 cm/sec 2 . 

(c) The light was flashing at a rate of jk = 29.75 flashes per second. 



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138 Chapter 3 Differentiation 

20. (a) 



(b) 



100 
50 



-50 
-100 

-150 
-200 






-50 



a 
20 



a«30-6t 



10 12 14 



21. C = position, A = velocity, and B = acceleration. Neither A nor C can be the derivative of B because B's 
derivative is constant. Graph C cannot be the derivative of A either, because A has some negative slopes while 
C has only positive values. So, C (being the derivative of neither A nor B) must be the graph of position. 
Curve C has both positive and negative slopes, so its derivative, the velocity, must be A and not B. That 
leaves B for acceleration. 

22. C = position, B = velocity, and A = acceleration. Curve C cannot be the derivative of either A or B because 
C has only negative values while both A and B have some positive slopes. So, C represents position. Curve C 
has no positive slopes, so its derivative, the velocity, must be B. That leaves A for acceleration. Indeed, A is 
negative where B has negative slopes and positive where B has positive slopes. 



23. (a) c(100) = 11,000 



11,000 
100 



$110 



(b) c(x) = 2000 + lOOx - .lx 2 =^ c'(x) = 100 - .2x. Marginal cost = c'(x) =4> the marginal cost of producing 100 
machines is c'( 100) = $80 

(c) The cost of producing the 101 s1 machine is c(101) - c(100) = 100 - ^ = $79.90 



24. (a) r(x) = 20000(1 - ±) 
(b) r'(100) = ?M» = $2. 



r'(x) 



20000 



which is marginal revenue. 



(c) lim r'(x) 



lim 



;o(iiio 



0. The increase in revenue as the number of items increases without bound 



will approach zero. 



25. b(t) = 10 G + 10 4 t - 10 3 t 2 => b'(t) = 10 4 - (2) (10 3 t) = 10 3 (10 - 2t) 

(a) b'(0) = 10 4 bacteria/fir (b) b'(5) = bacteria/hr 

(c) b'(10) = -10 4 bacteria/hr 

26. Q(t) = 200(30 - t) 2 = 200 (900 - 60t + t 2 ) => Q'(t) = 200(-60 + 2t) => Q'(10) = -8,000 gallons/min is the rate 
the water is running at the end of 10 min. Then Q j Q( = — 10,000 gallons/min is the average rate the 

water flows during the first 10 min. The negative signs indicate water is leaving the tank. 



Copyright (c) 2006 Pearson Education 




Section 3.3 The Derivative as a Rate of Change 139 



27. (a) y = 6(l-^) 2 = 6(l-i + ^) =* $ = £-1 

(b) The largest value of ^ is m/h when t = 12 and the fluid level is falling the slowest at that time. The 
smallest value of ^ is — 1 m/h, when t = 0, and the fluid level is falling the fastest at that time. 

the graph of y is 



dy 



(c) In this situation, ^ < 

always decreasing. As -^ increases in value, 
the slope of the graph of y increases from — 1 
to over the interval < t < 12. 



y 
















S 










5 










4 










3 




y- 


6(,- 


\2> 


1 






^ 












"12 




"dy~ 


t 








dt 


12 







28. (a) V = | Trr 3 =>• ^ = 4vrr 2 => ^| r _ 2 = 4tt(2) 2 = 16tt ft 3 /ft 

(b) When r = 2, ^ = 167T so that when r changes by 1 unit, we expect V to change by approximately \6ir. 
Therefore when r changes by 0.2 units V changes by approximately (167r)(0.2) = 3.2tt w 10.05 ft 3 . Note 
that V(2.2) - V(2) w 11.09 ft 3 . 



29. 200km/hr: 
t = 25, D = 



55 ^m/sec 
i?(25) 2 



500 



10 t 2 



6250 



4 m/sec, and D= y t 
m 



ft. ThusV 



500 
9 



20 



500 
9 



t = 25 sec. When 



30. s = v t - 16t 2 =>- v = v - 32t; v = =>• t = f§ ; 1900 = v t 
^ v = v/^OCl^OO) = 80^ ft/sec and, finally, ^vj[ft 



60 sec 
1 min 



16r so that t 

60 min 



1 hr 



1 mi 
5280 ft 



=> 1900 
238 mph. 



32 



64 



31. 





, 




600 




^ s = 20te-16r 2 


400 






200 


/ 4^ = 200 - 

/ 1 l^*"M 


-32< \ 


-200 


s 1 

- ^=-32 

d? 


>V "--^ 12 



body moves down 



(a) v = when t = 6.25 sec 

(b) v > when < t < 6.25 =>■ body moves up; v < when 6.25 < t < 12.5 

(c) body changes direction at t = 6.25 sec 

(d) body speeds up on (6.25, 12.5] and slows down on [0, 6.25) 

(e) The body is moving fastest at the endpoints t = and t = 12.5 when it is traveling 200 ft/sec. It's 
moving slowest at t = 6.25 when the speed is 0. 

(f) When t = 6.25 the body is s = 625 m from the origin and farthest away. 



Copyright (c) 2006 Pearson Education 




140 Chapter 3 Differentiation 

32. 



33. 




(a) v = when t = | sec 

(b) v < when < t < 1.5 =^ body moves down; v > when 1.5 < t < 5 =4> body moves up 

(c) body changes direction at t = | sec 

(d) body speeds up on (|, 5] and slows down on [0, |) 

(e) body is moving fastest at t = 5 when the speed = |v(5)| = 7 units/sec; it is moving slowest at 
t = | when the speed is 

(f) When t = 5 the body is s = 12 units from the origin and farthest away. 




(a) v = when t = ^f^- sec 

(b) v < when ^/^ < t < ^of^ => body moves left; v > when < t < ^/^ or ^f^- < t < 4 

=> body moves right 

(c) body changes direction at t = — f- — sec 

(d) body speeds up on (^^, 2) U (^±^> 4 1 and slows down on lo, ^^) U (2, ^4^) . 

(e) The body is moving fastest at t = and t = 4 when it is moving 7 units/sec and slowest at t = — f — se< 

(f) When t = ^j- — the body is at position s « —6.303 units and farthest from the origin. 



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Section 3.4 Derivatives of Trigonometric Functions 141 



34. 



10 




/V\ 


5 




/ s =4 - 7t 


-5 


/ 1 

/\ 


2\ 3 \ 4 




dt / + 


12t- 3t 2 \ 


-10 


d J 


. .... A 



+ 6t 2 -t 3 



- t 



—5 - 1 2 - 6t 



(a) v = when t = ^f^ 

(b) v < when < t < 6 ~f^ or 6 + f^ < t < 4 =4> body is moving left; v > when 

6-^/Tj < ( < 6+^As ^ body is moving right 

(c) body changes direction at t = — f- — sec 



(d) body speeds up on ( — f- — , 2 I 



U ( — ^ — 5 4 1 and slows down on |0, 



6-y/H 



)u( 2 , 6 ±#) 



(e) The body is moving fastest at 7 units/sec when t = and t = 4; it is moving slowest and stationary at 

6±-v/l5 



t 



3 



(f) When t = -^f- — the position is s w 10.303 units and the body is farthest from the origin. 



35. (a) It takes 135 seconds. 



AF 



(b) Average speed = ^- 



5 - 
73-0 



yij « 0.068 furlongs/sec. 



(c) Using a symmetric difference quotient, the horse's speed is approximately 4f - ,,,.. .,.. - — 



4-2 



0.077 furlongs/sec. 



(d) The horse is running the fastest during the last furlong (between the 9th and 10th furlong markers). This furlong takes 
only 1 1 seconds to run, which is the least amount of time for a furlong. 

(e) The horse accelerates the fastest during the first furlong (between markers and 1). 

3.4 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS 



1. y = — lOx + 3 cos x 

2. y = - + 5 sin x 

3. y = esc x — 4,/x + 7 =>• 



10 + 3 -f (cosx) = -10 -3 sin x 



- - # + 5£(sinx)= ^+5cosx 



dx 






-CSC x cot x — -A- + = —csc x cot X 

2,/x 



4. y = x 2 cot x 



^ dx 

-x 2 csc 2 x + 2x cot x 



x 2 £ (cot x) + cot x • i (x 2 ) + 4j = -x 2 csc 2 x + (cot x)(2x) 



(sec x + tan x) ^ (sec x — tan x) + (sec x — tan x) -r- (sec x + tan x) 



5. y = (sec x + tan x)(sec x — tan x) => -j| - ,..^ . .„,, , w Jn 

= (sec x + tan x) (sec x tan x — sec 2 x) + (sec x — tan x) (sec x tan x + sec 2 x) 

= (sec 2 x tan x + sec x tan 2 x — sec 3 x — sec 2 x tan x) + (sec 2 x tan x — sec x tan 2 x + sec 3 x — tan x sec 2 x) = 



( Note also that y = sec 2 x - tan 2 x = (tan 2 x + 1 ) - tan 2 x = 1 =4> ^ = 0. J 



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142 Chapter 3 Differentiation 



6. y = (sin x + cos x) sec x => g| = (sin x + cos x) -4 (sec x) + sec x -g (sin x + cos x) 

(sin x + cos x) sin x . cos x - sin x 



dx v ' dx 

(sin x + cos x)(sec x tan x) + (sec x)(cos x — sin x) 

sin 2 x + cos x sin x + cos 2 x — cos x sin x 1 



+ 



sec - x 



( Note also that y = sin x sec x + cos x sec x = tan x + 1 =>• i = sec 2 x. I 

7 v — cotx . dy _ (1 + cot x ) s ( cot x ) ~ (cot x > s (' + cot x ) _ (1 + cot x) (-esc 2 x) - (cot x) (-esc 2 x) 

' J ~ 1+cotx dx — (1+cotx) 2 (1+cotx) 2 

—esc 2 x — CSC 2 x cot X + CSC 2 x cot x — CSC 2 X 

— (1+cotx) 2 



(1 +cotx) 2 



_ cosx . dy _ (1 + sin x) ^ (cos x) - (cos x) ^ (1 + sin x) _ (1 + sin x) (-sin x) - (cos x) (cos x) 
* 1+sinx dx (1+sinx) 2 (1+sinx) 2 
— sin x — sin 2 x — cos 2 x — sin x — 1 —(1+sinx) —1 



(1+sinx) 2 (1+sinx) 2 (1+sinx) 2 1+sinx 



9. y = — — h — — = 4 sec x + cot x => -r = 4 sec x tan x — esc 2 x 

J cos x tan x dx 

10. y 



+ 



dy x(-sinx)-(cosx)(l) . (cos x)(l) - x(-sin x) _ -x sin x - cos x ■ cosx + x sin x 



x cos x dx x 2 cos 2 x x 2 cos 2 x 

11. y = x 2 sin x + 2x cos x — 2 sin x =S> ■£ = (x 2 cos x + (sin x)(2x)) + ((2x)(— sin x) + (cos x)(2)) — 2 cos x 

= x 2 cos x + 2x sin x — 2x sin x + 2 cos x — 2 cos x = x 2 cos x 

12. y = x 2 cos x — 2x sin x — 2 cos x =>• g| = (x 2 (— sin x) + (cos x)(2x)) — (2x cos x + (sin x)(2)) — 2(— sin x) 

= —x 2 sin x + 2x cos x — 2x cos x — 2 sin x + 2 sin x = — x 2 sin x 



13. s = tan t - t =4> f t = | (tan t) - 1 = sec 2 1 - 1 = tan 2 t 



14. s = t 2 - sec t + 1 =>• 21 = 2t - f (sec t) = 2t - sec t tan t 



-i c 1 +csc t , ds (1 — CSC t)(— esc t cot t) — (1 + CSC t)(csc t cot t) 

ID. S— 1 _ csct =? dt - (1-csct) 2 

—esc t cot t + CSC 2 t cot t — CSC t cot t — CSC 2 t cot t —2 CSC t cot t 

(1-csct) 2 



16. s 



(1 -CSC t) 2 

sint . ds _ (1 - cos t)(cos t) - (sin t)(sin t) cos t - cos 2 1 - sin 2 t _ cos t - 1 



1 — cos t dt 

1 



(1 -cos t) 2 



(1— cost) 2 (1— cost) 2 1 — COS t 



COS t — 1 



17. r = 4 - 2 sin =>■ % = - (0 2 j- g (sin 0) + (sin 0)(20)) = - (6» 2 cos 6 + 26 sin 9) = -6(6 cos 9 + 2 sin 9) 

18. r = 6 sin 6 + cos 6 => f g = (6 cos 9 + (sin 0)(1)) - sin 9 = 9 cos 6 

19. r = sec 6 esc 6 => % = (sec 0)(-csc 6 cot 0) + (esc 0)(sec tan 0) 

= (^i ) ( i ) (coi|) + ( i ) ( i ) (ml) = ^l + l = S ec 2 - esc 2 

V cos 8 J \sm 9 / \ sm9 J \ sm 9 / V cos # / V cos 9 / sin 2 # cos 2 9 

20. r = (1 + sec 0) sin =^ ^ = (1 + sec 0) cos + (sin 0)(sec tan 0) = (cos + 1) + tan 2 = cos + sec 2 t 

21 - p = 5 + ciq = 5 + tan q =► o| = sec2 q 



22. p = (1 + esc q) cos q => ^ = (1 + esc q)(— sin q) + (cos q)(— esc q cot q) = (—sin q — 1) — cot q = —sin q — esc q 

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Section 3.4 Derivatives of Trigonometric Functions 143 



23. p 



sin q + cos q 
cos q 



dp (cos q)(cos q — sin q) — (sin q + cos q)(— sin q) 



dq 



cos z q — cos q sin q + sin^ q + cos q sin q 1 J n 

ts 2 n ens 2 n VI 



cos z q 
cos 2 q 



r*A tan q _. dp (1 + tan q) (sec 2 q) — (tan q) (sec 2 q) sec 2 q + tan q sec 2 q — tan q sec 2 q sec 2 q 

P — 1 + tan q dq — (1 + tan q) 2 — (1 +tanq) 2 — (1 +tanq) 2 

25. (a) y = esc x => y' = —esc x cot x =4- y" = — ((esc x) (—esc 2 x) + (cot x)(— esc x cot x)) = esc 3 x + esc x cot 2 x 

= (esc x) (esc 2 x + cot 2 x) = (esc x) (esc 2 x + esc 2 x — 1) = 2 esc 3 x — esc x 
(b) y = sec x => y' = sec x tan x =>■ y" = (sec x) (sec 2 x) + (tan x)(sec x tan x) = sec 3 x + sec x tan 2 x 
= (sec x) (sec 2 x + tan 2 x) = (sec x) (sec 2 x + sec 2 x — 1) = 2 sec 3 x — sec x 

26. (a) y = —2 sin x =^ y' = —2 cos x =^ y" = — 2(— sin x) = 2 sin x => y'" = 2 cos x =4> y^ = —2 sin x 
(b) y = 9 cos x => y' = —9 sin x =$■ y" = —9 cos x =$■ y'" = — 9(— sin x) = 9 sin x =$■ y( 4 ' = 9 cos x 



27. y = sin x => y' = cos x =4- slope of tangent at 
x = — 7r is y'(— 7r) = cos (— 7r) = — 1; slope of 
tangent at x = is y'(0) = cos (0) = 1; and 
slope of tangent at x = ^ is y' (4^) = cos ^ 
= 0. The tangent at (— n, 0) is y — = — l(x + tt), 
or y = —x — 7r; the tangent at (0, 0) is 
y — = l(x — 0), or y = x; and the tangent at 
(f,-l)isy = -l. 




28. y = tan x =^ y' = sec 2 x =>- slope of tangent at x = — | 
is sec 2 (— | ) =4; slope of tangent at x = is sec 2 (0) = 1; 
and slope of tangent at x = | is sec 2 (|) =4. The tangent 

at (- f , tan(- f )) = (- f , -y^) is y + ^3 = 4(x + f ) 

the tangent at (0, 0) is y = x; and the tangent at (|, tan (|)) 

= (f, v / 3)isy-V / 3 = 4(x-f). 



y=.4x + f-VS 




y-x 



y - tan x 



y-4x-^ + V5 I 



29. y = sec x =4- y' = sec x tan x =>• slope of tangent at 

x = — | is sec (— |) tan (— |) = — 2y 3 ; slope of tangent 
at x = | is sec (|) tan (|) = \J 2 . The tangent at the point 
(- § , sec (- f )) = (- f , 2) is y - 2 = -2y/l (x+ |) ; 
the tangent at the point (f , sec ( ^)) = ( |, \/l\ is y - \fl 

= V^(x-l). 





< 


y = sec x 


' \ (-77/3, 2) 

! < 2 




/ 


! V <i 




(tt/4, •&) 


1 1 


y = ^2x 


4 


-7T/2 -7T/3 


ttIA 77/2 



slope of tangent at 



30. y = 1 + cos x => y' = —sin x 

x = — | is —sin ( — f ) = 2 ' s l°P e of tangent at x = y 
is —sin (y) — 1- The tangent at the point 

TT 3-\ 



(-f,l + cos (-!)) = (- f.f) 



is y — 



^3 



COS 



(¥)) 



; the tangent at the point 
^,1) isy-1 =x- f 




V3/ it\ 
'--2-( X + 3) 



(3k/2,1) 



Copyright (c) 2006 Pearson Education 




144 Chapter 3 Differentiation 

3 1 . Yes, y = x + sin x =>• y' = 1 + cos x; horizontal tangent occurs where 1 + cos x = =>• cos x = — 1 

=> X = 7T 



32. No, y = 2x + sin x => y' = 2 + cos x; horizontal tangent occurs where 2 + cos x = 
are no x-values for which cos x = —2. 



cos x = —2. But there 



33. No, y = x — cot x =>• y' = 1 + esc 2 x; horizontal tangent occurs where 1 + esc 2 x = 
are no x-values for which esc 2 x = — 1. 



1. But there 



34. Yes, y = x + 2 cos x =>• y' = 1 — 2 sin x; horizontal tangent occurs where 1 — 2 sin x = =>• 1 = 2 sin x 



i = sinx =>• x=Jorx=4r 

J DO 



35. We want all points on the curve where the tangent 
line has slope 2. Thus, y = tan x => y' = sec 2 x so 
that y' = 2 => sec 2 x = 2 =>■ sec x = ± \/2 
=^> x = ± |. Then the tangent line at (?, l) has 
equation y— 1=2 (x— |); the tangent line at 
(- f , -l) has equation y + 1 = 2 (x + f ) . 




36. We want all points on the curve y = cot x where 
the tangent line has slope — 1 . Thus y = cot x 

=> y' = —esc 2 x so that y' = — 1 =4> —esc 2 x = — 1 
=>■ esc 2 x = 1 => csc x= ±1 =>• x=|. The 
tangent line at (|, 0) is y = — x + | . 




37. y = 4 + cot x - 2 csc x =4> y' = -csc 2 x + 2 csc x cot x = - ( J-) C~ 2cosx ) 

^ ^ V sin x / V sin x / 

(a) When x = |, then y' = — 1; the tangent line is y = — x + | + 2. 

(b) To find the location of the horizontal tangent set y' = => 1—2 cos x = 
then y = 4 — y 3 is the horizontal tangent. 



x = | radians. When x = |, 



38. y = 1 + \/2cscx + cotx => y' = - y^csc x cot x - csc 2 x = - (J^) ( y ^f ±1 ) 
(a) If x = j, then y' = —4; the tangent line is y = — 4x + n + 4. 



(b) To find the location of the horizontal tangent set y' = 



'2 cos x + 1 = 



x = 2s radians. When 



x = ; ?, then y = 2 is the horizontal tangent. 



39. lim sin (i - i) = sin (| - |) = sin = 



40. lim ^ y 7 ! + cos(7r csc x) = </l + cos (7r csc (— f)) = yl + cos (w • (—2)) = y 2 



41. lim sec [cos x + 7r tan ( 4 s ^ c - ) — 1] = sec [cos + ir tan ( 4 — Q ) — l] = sec [l + w tan (|j 



— 11= sec 7T = — 1 



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Section 3.4 Derivatives of Trigonometric Functions 145 



42. lim sin ( t ?r+t ™ x ) = sin ( , 7r +^"° a ) 

x j, q V tan x— 2 sec x / V tan 0—2 sec / 



43. 



lim tan (l - sal) = tan (l - lim ^ = tan(l - 1) = 
t-»o v ' ' V t->0 l / 



44. lim cos (4^ 

-> V Sln e 



) = COS I 7T lim -/-a ) = COS I 7T • ^TTT ) = cos (w - t) = — 1 



45. s = 2 - 2 sin t => v = ^ 

at 



-2 cos t =>• a = ^ = 2 sin t =^>j=^ = 2 cos t. Therefore, velocity = v (|) 



- V 2 m/sec; speed = |v (|) | = y 2 m/sec; acceleration = a (|) = y 2 m/sec 2 ; jerk = j (?) = y2 m/sec 3 . 



46. s = sin t + cos t =>• v = ^ = cos t — sin t =>■ a = $ 

at at 



da 



-sin t — COS t =>- J — h 

velocity = v (? ) =0 m/sec; speed = |v (?) | =0 m/sec; acceleration = a (?) = — y 2 m/sec 2 ; 
jerk = j (|) = m/sec 3 . 



cos t + sin t. Therefore 



47. lim A f(x) = lim n s -^ = lim n 9 (^) (^) = 9 so that f is continuous at x = =>• lim n f(x) = f(0) 



x -> x^ 

=> 9 = c. 



x^O 



x^O 



lim_ g(x) = lim_ (x + b) = b and lim g(x) = lim cos x = 1 so that g is continuous at x = =>• lim_ g(x) 

x^O x^O x^0 + x ^ 0+ " x^O 

= lim g(x) => b = 1. Now g is not differentiable at x = 0: At x = 0, the left-hand derivative is 

x — > 0+ 

^ (x + b)| = 1, but the right-hand derivative is -^ (cos x)| _ = — sin = 0. The left- and right-hand 
derivatives can never agree at x = 0, so g is not differentiable at x = for any value of b (including b = 1). 



g^ggg (cos x) = sin x because f^ (cos x) = cos x => the derivative of cos x any number of times that is a 



multiple of 4 is cos x. Thus, dividing 999 by 4 gives 999 = 249 -4 + 3 



d\"" 



(cos x) 



£s ;|ss!je (cos x) = £s (cos x) = sin x. 



50. (a) y = sec x 



l 

COS X 



dy _ (cosx)(0)-(l)(-sinx) 



dx 



(cos x) 2 



=> 4- (sec x) = sec x tan x 

(b) y = esc x 
d 



l 

sin x 



dy _ (sinx)(0)-(l)(cos x) _ -cos x 



dx 



(sin x) 2 



L = (i)(M) 
x V cos x / V cos x / 

/ cos X \ 
V sin x / 



' -1 \ Ccosx' 
v sin x / 



sec x tan x 



-esc x cot x 



=> ^ (esc x) = —CSC x cot X 



(c) y = cot x 



cos X 
sin x 



dy (sin x)(— sin x) — (cos x)(cos x) 

dx (sin x) 2 



— sin x— cos x —1 



=>• 4- (cot x) = —CSC 2 X 



51. 





As h takes on the values of 1, 0.5, 0.3 and 0. 1 the corresponding dashed curves of y = Sln (x ± J — !HL5 g et 

closer and closer to the black curve y = cos x because 4- (sin x) = lim Mn . Sln x = cos x. The same 

} dx h - h 

is true as h takes on the values of —1, —0.5, —0.3 and —0.1. 



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146 Chapter 3 Differentiation 



52. 





As h takes on the values of 1, 0.5, 0.3, and 0.1 the corresponding dashed curves of y 



cos (x + h) — cos x 



get 



closer and closer to the black curve y = — sin x because 4- (cos x) = lim co , — ^^ = —sin x. The 

3 dx h - h 

same is true as h takes on the values of — 1, —0.5, —0.3, and —0.1. 



53. (a) 





£ 


-\ 




/f- 




A 5 


\ 




/ 


-7T 

ij 


-0.5 




\ n 1 


2jr 


-J 


-1 




\^/ 





The dashed curves of y 



sin(x + h) — sin(x — h) 
2h 



are closer to the black curve y = cos x than the corresponding dashed 



curves in Exercise 5 1 illustrating that the centered difference quotient is a better approximation of the derivative of 
this function. 



(b) 





/°=\ X 














r \ 






/' 


~\\ 




/ 


o\s 






/ 




\ 


-7T 


-0.5 

-1 


\ 


/ 






27T 



The dashed curves of y = — > ; 2h -^ ' are closer to the black curve y = — sin x than the corresponding dashed 

curves in Exercise 52 illustrating that the centered difference quotient is a better approximation of the derivative of 
this function. 



54. lim 
h^0 



|0 + hl-|0-h| 



lim 

x^0 



- h| 



lim 







2h x _, 2h 

though the derivative of f(x) = |x| does not exist at x 



the limits of the centered difference quotient exists even 
= 0. 



55. y = tan x => y' = sec 2 x, so the smallest value 
y' = sec 2 x takes on is y' = 1 when x = 0; 
y' has no maximum value since sec 2 x has no 
largest value on (— |, |) ; y' is never negative 
since sec 2 x > 1. 




Copyright (c) 2006 Pearson Education 




Section 3.4 Derivatives of Trigonometric Functions 147 



56. y = cot x =>- y' = —esc 2 x so y' has no smallest 
value since —esc 2 x has no minimum value on 
(0, 7r); the largest value of y' is — 1, when x = | ; 
the slope is never positive since the largest 
value y' = —esc 2 x takes on is — 1. 




57. y - Mn v 



lim 5»x 

x^O x 



at y 



— appears to cross the y-axis at y = 1, since 

1 ; y = ^^ appears to cross the y-axis 
2, since lim 5M>2x = 2; y = sin4x 







appears to 



cross the y-axis at y = 4, since lim 







However, none of these graphs actually cross the y-axis 
since x = is not in the domain of the functions. Also, 



lim ^^ 



5, lim 



sin(— 3x) 



-3, and lim ^ 

x^ ' 
sin(— 3x) 



kx 



= k => the graphs of y = ^5, y = >^-^>, and 
y = ^^ approach 5, —3, and k, respectively, as 
x — > 0. However, the graphs do not actually cross the 
y-axis. 




y ■ (sin 4x)/x 

y = (sin 2x)/x 
y = (sinx)/x 



58. (a) h 


sin h 
h 


(^) (^) 


1 


.017452406 


.99994923 


0.01 


.017453292 


1 


0.001 


.017453292 


1 


0.0001 


.017453292 


1 



lim 



lim 



>( h -ii) 



lim 
h->0 



'("-Tin) 






lim ^-, 



ISO 



(converting to radians) 



h 


cos h— 1 
h 


1 


-0.0001523 


0.01 


-0.0000015 


0.001 


-0.0000001 


0.0001 






lim £2^" 
h^O h 



0, whether h is measured in degrees or radians. 



(0 = h • Tin) 



(c) In degrees, -jj (sin x) 
= lim (sin x • ^ 

u . n V h 



lim 



sin (x + h) — sin x 



•0 
(sinx)(0) + (cos x) (t| 



+ lim (cos x 



i- (sin x cos h + cos x sin h) — sin x 

sin h ^ (c\ n v \ U-m /"cosh—i 



(sin x) • lim 
twO 



(cosx)- lim m±\ 



180/ 



COS X 



(d) In degrees, 4- (cos x) = lim 

W B dx V > h _> Q 



" 180 

cos (x + h) — cos x 
h 



lim 



(cos x cos h — sin x sin h) — cos x 



lim 



(cos x)(cos h — 1) — sin x sin h 



(cos x) lim 
h->0 



(e) ^(sinx) 



_d_ 

dx V 180 



h — n 

- lim (cosx- £2iA^i) - lim (sinx-^) 
h h^O v h > h->0 v h ; 

£2^1) - (sin x) h lim o (^) = (cos x)(0) - (sin x) (^) = 

d_ 

dx 



180 



cos x; 



cosx ) = - (tm) 2 sin x ; a& ( sin x > = ^ (- (ifo) 2 sin x ) = -(ifo) ; 

£ (cos x) = £ (- jfg sin x) = - (yffj) 2 cos x; £, (cos x) = £ (- (^f cos x) = {^f sin x 



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148 Chapter 3 Differentiation 

3.5 THE CHAIN RULE AND PARAMETRIC EQUATIONS 

1. f(u) = 6u - 9 => f (u) = 6 => f (g(x)) = 6; g(x) = ^x 4 ^ g'(x) = 2x 3 ; therefore g = f (g(x))g'(x) 

= 6 • 2x 3 = 12x 3 

2. f(u) = 2u 3 => f (u) = 6u 2 => f (g(x)) = 6(8x - l) 2 ; g(x) = 8x - 1 => g'(x) = 8; therefore g = f'(g(x))g'(x) 

= 6(8x- l) 2 -8 = 48(8x- l) 2 

3. f(u) = sin u =>- f'(u) = cos u => f'(g(x)) = cos (3x + 1); g(x) = 3x + 1 =>• g'(x) = 3; therefore g = f'(g(x))g'(x) 

= (cos (3x + 1))(3) = 3 cos (3x + 1) 



4. f(u) = cos u => f'(u) = -sin u => f'(g(x)) = -sin (^) ; g(x) = f => g'(x) = - §; therefore | = f'(g(x))g'(x) 



-sin 



(f)-(f) = ^n(f) 



5. f(u) = cos u => f'(u) = —sin u =$■ f'(g(x)) = —sin (sin x); g(x) = sin x =>- g'(x) = cos x; therefore 



% = f'(g(x))g'(x) = -(sin (sin x)) cos x 



6. f(u) = sin u =>- f'(u) = cos u => f'(g(x)) = cos (x — cos x); g(x) = x — cos x => g'(x) = 1 + sin x; therefore 

dy 
dx 



^ - f'(g(x))g'(x) = (cos(x - cos x))(l + sin x) 



7. f(u) = tan u => f'(u) = sec 2 u =4> f'(g(x)) = sec 2 (lOx - 5); g(x) = lOx - 5 => g'(x) = 10; therefore 



g = f'(g(x))g'(x) = (sec 2 (lOx - 5)) (10) = 10 sec 2 (lOx - 5) 



]. f(u) = —sec u =4> f '(u) = —sec u tan u =>• f'(g(x)) = —sec (x 2 + 7x) tan (x 2 + 7x) ; g(x) = x 2 + 7x 

dy 
dx 



g'(x) = 2x + 7; therefore g = f'(g(x))g'(x) = -(2x + 7) sec (x 2 + 7x) tan (x 2 + 7x) 



9. With u = (2x + 1), y = u 5 : g = g g = 5u 4 • 2 = 10(2x + l) 4 



10. With u = (4 - 3x), y = u 9 : g = g g = 9u 8 • (-3) = -27(4 - 3x) 8 



11. Withu=(l-f),y = ^: g = ^g = -7u- 8 -(-i) = (l-f)- 



12. Withu= f|-l),y = u- 10 : g = £ £ = -lOir 11 - '* 



v 2 / ' ' dx du dx 

13. Withu= (f +x-i),y 



u 



4. dy _ dy du 
dx du dx 



2 ) = 5(| - 1) ^ 

4u 3 -(t + l + ^)=4(f+x-i) 3 (| + l + i) 



14. Withu=(f + i),y = ^:g = g£=5u 4 .(I-^) = (f + i) 4 (l-^ 

15. With u = tan x, y = sec u: gj = gg gj = ( sec u tan u) (sec 2 x) = (sec (tan x) tan (tan x)) sec 2 x 

16. Withu = 7 r-i,y = cotu: g = £g = (-esc 2 u) (4,) = - 4, esc 2 (» - 1) 

17. With u = sin x, y = u 3 : g = g f = 3u 2 cos x = 3 (sin 2 x) (cos x) 



18. With u = cos x, y = 5tT 4 : g = g g = (-20iT 5 ) (-sin x) = 20 (cos~ 5 x) (sin x) 



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Section 3.5 The Chain Rule and Parametric Equations 149 



19. p = y/3=l=(?-t)V* =* ^ = I(3-tr 1 /2. d ( 3_ t) = _i (3 _ t) -i/2 



2V3-t 



20. q = i/2r - r 2 = (2r - r 2 ) 1/2 => f = \ (2r - r 2 ) _1/2 - | (2r - r 2 ) = \ (2r - r 2 ) _1/2 (2 - 2r) 



^ 



21. s = #■ sin 3t + #- cos 5t => jj? = 4- cos 3t • £ (3t) + #■ (-sin 5t) • £ (5t) = ^ cos 3t - ^ sin 5t 

3tt 5tt at in at v 7 5tt v ' at v ' tt tt 

= - (cos 3t — sin 5t) 



22. s = sin(^) +cos(^) 



'37rt\ ds 



COS 



m ■!(¥)- •!» (¥) ■ I (¥) = ¥ cos (f ) - f sin (f ) 



f (cos 3f£ - sin 2si) 



cot 9) 2 (esc 9 + cot 9) 2 



23. r = (esc + cot 6y l => % = -(esc 9 + cot 0)~ 2 ^ (esc + cot (9) = ™J> «* <> + <*<? o - ^ '-' --^ ^ ' "-' 

CSC 9 

CSC 9 + cot 9 

sec 9 tan 9 + sec 2 9 _ sec 9 (tan 9 + sec 9 

tan 9) 2 ~~ (sec 9 + tan 9) 2 



24. r = -(sec 9 + tan 9)~ x ^ % = (sec + tan 0)-' 2 ^ (sec + tan 0) = se ( c s f c 7 + 6 



25. y = x 2 sin 4 x + x cos -2 x =^> g = x 2 £■ (sin 4 x) + sin 4 x • ^ (x 2 ) + x I- (cos -2 x) + cos -2 x • ^ (x) 

= x 2 (4 sin 3 x -£ (sin x)) + 2x sin 4 x + x (—2 cos -3 x • ^ (cos x)) + cos -2 x 
= x 2 (4 sin 3 x cos x) + 2x sin 4 x + x( (—2 cos -3 x) (—sin x)) + cos -2 x 
= 4x 2 sin 3 x cos x + 2x sin 4 x + 2x sin x cos -3 x + cos -2 x 

26. y = - sin -5 x — | cos 3 x =>- y' = - £ (sin -5 x) + sin -5 x-r(-)-|i (cos 3 x) - cos 3 x - -jj- (|) 

= i (-5 sin~ 6 x cos x) + (sin -5 x) (— 4J — | ( (3 cos 2 x) (-sin x)) - (cos 3 x) (|) 
= — - sin -6 x cos x — 4 sin -5 x + x cos 2 x sin x — \ cos 3 x 

x x 2 3 



27. y= ^(3x-2V + (4- 5 i 2 )- 1 => g = ^(3x-2)e.A ( 3x-2) + (-l)(4-^)- 2 .A( 4 -^) 
= ^(3x-2) 6 .3 + (-l)(4-^)- 2 (i)=(3x-2) 6 



x^ 4- 1 



28. y = (5 - 2x) 

6 

(5-2x)4 



- 3 + H= 
+o 3 



dy 
dx 



-3(5 - 2x)- 4 (-2) + i (| + 1) J (- |) = 6(5 - 2x)- 4 - (A) (§ 



29. y = (4x + 3) 4 (x + l) -3 => g = (4x + 3) 4 (-3)(x + 1)~ 4 - £ (x + 1) + (x + ir 3 (4)(4x + 3) 3 • £ (4x + 3) 
= (4x + 3) 4 (-3)(x + 1)" 4 (1) + (x + l)- 3 (4)(4x + 3) 3 (4) = -3(4x + 3) 4 (x + l)" 4 + 16(4x + 3) 3 (x + l)" 3 

= W? [" 3 ( 4x + 3 ) + 16 < x + !)1 = ( % 3 + ( i)* + 7) 



30. y = (2x - 5)- 1 (x 2 - 5x) G => g = (2x - 5r 4 (6) (x 2 - 5x) 5 (2x - 5) + (x 2 - 5x)°(-l)(2x - 5)" 2 (2) 



6 (x 2 - 5x 



5 _ 2(x 2 -5x) tl 
1 (2x - 5)2 



31. h(x) = x tan (2y^x) + 7 => h'(x) = x ^ (tan (2X 1 / 2 )) + tan (2X 1 / 2 ) - £ (x) + 

= x sec 2 (2X 1 / 2 ) • £ (2X 1 / 2 ) + tan (2X 1 / 2 ) = x sec 2 (2^/x) • 4- + tan (2^/x) = ,/x sec 2 (2-^/x) + tan (2,/x) 



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150 Chapter 3 Differentiation 

32. k(x) = x 2 sec (I) =* k'(x) = x 2 £ (sec 1) + sec (I) ■ A. ( x 2) = x 2 sec (I) tan (I) . A. (I) + 2x sec (V 

- ¥) + 2x sec (!) = 2x sec (!) ~ sec (l) tan (I) 



33- fW=(if^) ^f'W = 2( T f^)-^ u+cos 



d_ f sin > 
W 

(2 sin g) (cos g + cos 2 9 + sin 2 g) _ (2 sin 9) (cos 9+ 1) _ 2 sin 9 



2 sin 9 (1 + cos 9)(cos 9) - (sin g)(-sin I 
l+cos9 ' (l+cos9) 2 



(l+cos9) 3 



(l+cos9) 3 (1+c 



1 + cosf\~l 



34. g(t)=( i ^r => g'ct) = - (4^) 

_ — (— sin 2 t — cos t — cos 2 t) _ 1 

(1 +cost) 2 ~ 1 + cos t 



1 + cos t \ ^ d ( 1 + cos t 



dt V sin 



P)=- 



n 2 1 (sin t)(— sin t) — (1 + cos t)(cos t) 



( 1 + cos t) 2 



(sin t) 2 



35. r = sin (<9 2 ) cos (2d) => % = sin (d 2 ) (-sin 2d) j- g (2d) + cos (2d) (cos (<9 2 )) • ± ((9 2 ) 

= sin (0 2 ) (-sin 20)(2) + (cos 20) (cos (0 2 )) (20) = -2 sin (0 2 ) sin (20) + 2d cos (2(9) cos (0 2 ) 



36. r = (sec v^) tan (I) => % = (sec y/o) ( sec 2 i) (- £) + tan (J) (sec yfd tan ^) (^) 
= - ^ sec v^ sec 2 (\) + -^ tan (i) sec v^ tan y^ = (sec y^ 



tan ^0 tan (1) sec 2 (1) 
275 52_ 



37- q - * fe)^ = cc (^) - | (^) = cos (^) - ^*<^ 
= - (v^T) • ^^ = cos (^) (t^g) = (^) cos (-^) 
> g = -esc 2 (*1) - I (*«) = (-esc 2 (*!)) (ts-^M-l) 



38. q = cot(^ 



i 



39. y = sin 2 (?rt - 2) => f = 2 sin (vrt - 2) • f t sin (vrt - 2) = 2 sin (Trt - 2) • cos (vrt - 2) • | (71I - 2) 
= 27r sin (irt — 2) cos (7rt — 2) 



40. y = sec 



= sec 2 7rt =^ -| = (2 sec 7rt) - ^ (sec 7rt) = (2 sec 7rt)(sec 7rt tan 7rt) • 4 (7rt) = 27r sec 2 7rt tan 7rt 

2t)- 4 =^> I = -4(1 + cos 2tr 5 • I (1 + cos 2t) = -4(1 + cos 2t)~ 5 (-sin 2t) - | (2t) = ^ 

i)) = -2(l + cot(i))- 3 .(-csc 



42. y=(l+ C ot(§)) 

9 /• t N 



-liiL 



4 => | = -4(l+cos2tr 5 -^i 

| = -2(l + cot(i))- 3 .^( 



41. y = (1 +cos 

;i))- 3 -^(i + cot(|)) 



8 sin 2t 

~~ (1 +cos2t) 5 



{■2) J ' dt I 2 / 



(l+cot(|)) 3 

43. y = sin (cos (2t - 5)) => ^ = cos (cos (2t - 5)) - | cos (2t - 5) = cos (cos (2t - 5)) - (-sin 
= -2 cos (cos (2t - 5))(sin (2t - 5)) 



(2t - 5)) - A (2t - 5) 



; 5 sin(i)) 



44. y = cos 

i sin (I)) (cos (I 



dy 



(5 sin (I))- | (5 sin (|)) = 



-sin(5sin(i))(5cos(i))-^(i) 



— I sin (5 s 

^ = 3[l + tanM^)] 2 ^[l+tan4(^)]=3[l+tanM^)] 2 [4tan3( 1 L).d tan( ^ )] 
»' (T2) sec2 (12) - T2] = [1 + ^ (T2)] 2 K (T2) sec2 (&)] 



45. y=[l+tan^); 

(7t)f-2cos(7t)(-sin(7t))(7) 



= 12[l+tan 4 ( 1 L)] 2 [tan3( 1 L)sec 2 ( I L) 

I [1 + cos 



46. y = \ [1 + cos 2 (7t)f 



dy 



dt 6 I 



tan4 (ll)] • dt L 1 T lim IT2JJ =J [' 

• 4 (i> 3 (^ 

-7 [1 + cos 2 (7t)] 2 (cos (7t) sin (7t)) 



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Section 3.5 The Chain Rule and Parametric Equations 151 



47. y = (1 +cos(t 2 )) 



1/2 



at 



t 2-n-i/2 d 



2\\-V 2 t 



\ (1 + cos (t*))- 1 " • i (1 + cos (t 2 )) = I (1 + cos (f))~^ (-sin (t 2 ) • g (t 2 )) 



i(l + cos(t 2 ))- 1/2 (sin(t 2 ))-2t=-^ii 



^/l + cosCt 2 ) 



48. y = 4sin(\/l - v m] , £ = 4 cos ( ^/ 1 + ^ ) - £ ( ^1 + 0) = 4 cos ( yj\ + ) • ^-^ • £ (l + 0) 



2 cos 



(v^+\A) cos^i + yi^) 



v / iT7t-2v / t \A+^t 



49. y =(l + I)-=> y ' = 3(l + ir(-i)=-| r (l + ir=>y=(- JO" £(1 + D -(1 + D -£ (JO 

= ^ (i + i) (i + 1) 



50. y=(l- x /x)" 1 =► y ' = -(l-y x )" 2 (-i X - 1 /2) = Kl-^-V 1 ^ 

=► y« = i [(1 - ^)- 2 (- Ix-3/ 2 ) +x-i/2(_2) (1 - ^)- 3 (- |x-V2)] 

= i [f x-3/ 2 (i - y*-)- 2 + x-i (i - ,A)- 3 ] = i x -i (i - .A)- 3 [- |x-v2 (1 _ ^) + 1] 

= i(l-AA)- 3 (-^ + ^l)=i(l-^- 3 (|-^) 

51. y = icot(3x- 1) => y' = -icsc 2 (3x- 1)(3) = -|csc 2 (3x- 1) =>■ y" = (-f)(csc(3x- 1)- ^csc(3x- 1)) 

= -f csc(3x- l)(-csc(3x- l)cot(3x- 1)- £(3x- 1)) = 2 csc 2 (3x - l)cot(3x- 1) 

52. y = 9 tan (f) => y' = 9 (sec 2 (f )) (±) = 3 sec 2 (f) => y" = 3 - 2 sec (f ) (sec (f) tan (f )) (|) = 2 sec 2 (f ) tan (|) 

53. g(x) = y^ ={► g'(x) = i => g(l) = 1 and g'(l) = \ ; f(u) = u 5 + 1 => f (u) = 5u 4 =* f (g(l)) = f (1) = 5; 



therefore, (f o g)'(l) = f (g(l)) • g'(l) = 5 • 



2 2 



54. g(x) = (1 - x)- 1 =* g'(x) = -(1 - x)- 2 (-l) = ^ =* g(-l) = i and g'(-l) = i ; f(u) = 1 - 1 

=> f'(u)=i => f'(g(-l)) = f (|) =4; therefore, (fog)'(-l) = f'(g(-l))g'(-l) = 4-i = l 

55. g(x) = 5^ => g'(x) = jjj => g(l) = 5 and g'(l) = f ; f(u) = cot (f ) => f'(u) = -esc 2 (f ) (§) 



=2 csc 2 fZLH^ 
10 csl " WO/ 



f'(g(l)) = f'(5) 



is esc 2 (f ; 



10 



; therefore, (f o g)'(l) = f'(g(l))g'(l) = - * 



10 2 



56. g(x) = 7rx => g'(x) = 7r =^> g (|) = | and g' Q) = ir; f(u) = u + sec 2 u =>• f'(u) =1+2 sec u • sec u tan u 

= 1+2 sec 2 u tan u => f'(g (1)) = f (|) = 1 + 2 sec 2 | tan f = 5; therefore, (fog)' (1) = f'(g (1)) g' (I) = 5^ 



57. g(x) = 10x 2 + x + 1 => g'(x) = 20x + 1 => g(0) = 1 and g'(0) = 1; f(u) 



2u 
u 2 +l 



f'(u) 



(u 2 + l)(2)-(2u)(2u) 
(u2 + l) 2 



=f^±| =* f'(g(0)) = f'(l) = 0; therefore, (f o g)'(0) = f'(g(0))g'(0) = 0-1=0 



58. g(x)=i-l => g'(x) = -2 => g(-l) = 0andg'(-l) = 2;f(u)=(^j) 2 =* f'(u) = 2 (^j) * (^j) 

= 2 (£*) ■ (n + T + "i ( r lxl) = 2 i^W = £w =► f '(g(-D) = f '(°) = - 4 = therefore ' 

(f og)'(-l) = f'(g(-l))g'(-l) = (-4)(2) = -8 



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152 Chapter 3 Differentiation 



59. (a) y = 2f(x) =* g = 2f'(x) =* gl 



2f'(2) = 2(i) 



(b) y = f(x) + g(x) => | = f (x) + g'(x) => | = f (3) + g'(3) = 2^ + 5 



(c) y = f(x)-g(x) 



g = f(x)g'(x) + g(x)f (x) 



dy 
dx 



f(3)g'(3) + g(3)f (3) = 3 • 5 + (-4)(2tt) = 15 - 8tt 



(d) y = fJ r\ 

v ' J g(x) 



dy _ g(x)f'(x)-f(x)g'(x) 
dx [g(x)]2 



dy 
dx| , 

dy I 



g(2)f'(2)-f(2)g'(2) _ (2)(l)-(8)(-3) _ 37 
[g(2)p 22 6 



(e) y = f(g(x)) => g = f (g(x))g'(x) -- dx 

(f) y = (f(x))V2 ^ I = 1 (f(x))-i/a . f ( X ) 



f (g(2))g'(2) = f (2)(-3) = i (-3) = -1 



f'(x) 



dy 
dx 



f'(2) 



(i) 



sfl 



dx 2^^^ *w 2 yftx> "^ dx l x=2 2 \/f(2J 2x/8 6\/8 12-/2 24 

(g) Y = (gWr 2 => g = -2(g(x))- 3 • g'(x) => 1 1 = -2(g(3))- 3 g'(3) = -2(-4)- 3 • 5 



_5_ 
32 



(h) y = ((f(x)) 2 + (g(x)) 2 ) 1/2 



g = i ((f(x)) 2 + (g(x)) 2 ) 1/2 (2f(x) - f (x) + 2g(x) • g'(x)) 



dy 
dx 



->2\-l/2 



I x=2 

5 

3\/l7 



1 ((f(2)) 2 + (g(2)) 2 ) ^ (2f(2)f (2) + 2g(2)g'(2)) 



I Cq2 

2 



,2N-l/2 



+ 2 2 ) i/2 (2-8-i+2-2-(-3)) 



60. (a) y = 5f(x)-g(x) 



g = 5f (x) - g'(x) => g 



5f'(l)-g'(l) = 5(-i)-(f) = l 



(b) y = f(x)(g(x)) 3 => g = f(x) (3(g(x)) 2 g'(x)) + (g(x)) 3 f (x) =* gl = 3f(0)(g(0)) 2 g'(0) + (g(0)) 3 f (0) 



3(l)(l) 2 (i)+(l) 3 (5) = 6 



(c) y 



f(x) 



dy _ (g(x)+l)f'(x)-f(x)g'(x) . dy 



g(x)+l ^ dx (g(x)+l) 2 

( -4+l)(-l)-(3)(-|) _ 
(-4+1)2 



dx 



(g(l)+l)f'(l)-f(l)g'(l) 
(g(l) + l) 2 



(d) y = f(g(x)) 



dy 



f (g(x))g'(x) ={► g = f'(g(0))g'(0) = f'(l) Q) = (- |) (!)=-£ 



(e) y = g(f(x)) =* g = g'(f(x))f (x) =► gl = g'(f(0))f (0) = g'(l)(5) 



40 
3 



(f) y=(x n +f(x)) 2 => g = -2(x n +f(x)) 3 (llx 10 + f'(x)) 



dy 
dx 



I) (5) 
-2(l + f(l))- 3 (ll+f'(l)) 



-2(i + 3r3(ii-i) = (-| I )(f) 



(g) y = f(x + g(x)) => g = f (x + g(x)) (1 + g'(x)) => g 



f'(0 + g(0))(l + g'(0))=f'(l)(l + i) 



l)(!) 



61- £ = &■£: s = cose => § = -sin< 



-sin(f ) = 1 so that | 



dfl dt 



1-5 = 5 



62- g = g-£:y = * 



= x 2 + 7x-5 => g = 2x + 7 



Si =9sothat| = g.|=9-i=3 



dx 



x=l 



= x, we should get ^ = 1 for both (a) and (b): 

£ = 5; therefore, g 

dx ' ' dx 



63. Withy ,,.. Jx . , ..... 

(a) y = f + 7 => g = i ; u = 5x - 35 => | = 5; therefore, g = g-g = j-5 = l,as expected 

r 1 => g = -<* - i)- 2 (D = o^V : therefore 1 = 1-1 

/_ i\2 l 1 ■ * 



du — 5 ' "" -'-' "^ 

Cb) Y = i + ^ =* g = -^;u = (x-l)- 

= -r ■ , ~\v = ^ — -1 ,.2 • , "l,, = (x — l) 2 • -. — ^ = 1, again as expected 

U 2 (X— l) 2 ((x-1) -1 ) (x-l)2 V ' (X— 1)2 ' ° r 

With y = x 3 / 2 , we should get g = § x 1 / 2 for both (a) and (b): 

= U 3^ g =3u 2. u=v /^ | = _^ ;therefore) g = g.du = 3u2 ._^ = 3 ( ^ )2 ._^ = |^ 



(a) y 



as expected. 

dZ = 1 ; U = X 3 
du 2^/u 

again as expected. 



(b) y = s/u 



f = 3x 2 ; therefore, g = g ■ £ = i 

dx dx du dx 2a/ u 



dy du _ _J_ o 2 _ 1 
dx 2^ 2V5 



_ 1 o v 2 _ 3 1/2 

— n ATT JA ~~ 9 A 



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Section 3.5 The Chain Rule and Parametric Equations 153 



dx 



(2 sec 2 f) 



E af . n 2 7TX 
2 scl ' 4 



65. y = 2tan(f) 

= | sec 2 (|) = 7r =^ slope of tangent is 2; thus, y(l) = 2 tan (|) 
given by y — 2 = 7r(x — 1) =>■ y = 7rx + 2 — 7r 



(a) I 



2 and y'(l) 



tangent line is 



(b) y' = | sec 2 (^) and the smallest value the secant function can have in— 2<x<2isl =4> the minimum 
value of y' is § and that occurs when § = § sec 2 (™) =^ 1 = sec 2 (™) => ± 1 = sec (™) =4> x = 0. 



66. (a) y = sin 2x => y' = 2 cos 2x =4> y'(0) = 2 cos (0) = 2 =£- tangent to y = sin 2x at the origin is y = 2x; 

y = —sin (|) =>■ y' = — \ cos (|) =>■ y'(0) = — \ cos = — \ =^ tangent to y = —sin (|) at the origin is 
y = — \ x. The tangents are perpendicular to each other at the origin since the product of their slopes is 
1. 

(b) y = sin(mx) =4> y' = mcos(mx) =4> y'(0) = mcosO = m; y = —sin ( — ) => y' = — — cos (-) 

=> y'(0) = — — cos (0) = — — . Since m -(——)= —1, the tangent lines are perpendicular at the origin. 

(c) y = sin(mx) => y' = mcos(mx). The largest value cos (mx) can attain is 1 at x = =>- the largest value 
y' can attain is |m| because |y'j = |m cos (mx)| = |m| |cos mx| < |m| • 1 = |m| . Also, y = —sin (-) 

=> y' = — — cos (-) =$■ ly'l = I— cos (— ) I < I — I Icos ( — ) I < A => the largest value y' can attain is I — I . 

J m Vm/ I-' I Im Vm/I — Imll Vm/I — |m| to J I ml 

(d) y = sin (mx) =4> y' = m cos (mx) =4> y'(0) = m =$■ slope of curve at the origin is m. Also, sin (mx) completes 
m periods on [0, 27r]. Therefore the slope of the curve y = sin (mx) at the origin is the same as the number 

of periods it completes on [0, 2tt]. In particular, for large m, we can think of "compressing" the graph of 
y = sin x horizontally which gives more periods completed on [0, 2ir], but also increases the slope of the 
graph at the origin. 



67. x = cos 2t, y = sin 2t, < t < 7r 

=> cos 2 2t + sin 2 2t = 1 => x 2 + y 2 = 1 



68. x = cos (n — t), y = sin (n — t), < t < tt 
=^ cos 2 (V - t) + sin 2 (tt - t) = 1 
=> x 2 + y 2 = l,y > 





y 


















2 




* 2 + /=l 




l = irllf 






\(=0 




-2 -\\ 




-I 

-2 




/' 2 





2 2 

x + y -1 



1 .J- n/2 



69. x = 4 cos t, y = 2 sin t, < t < 2n 



16 cos 2 1 i 4sin 2 t 



16 



1 



16 



70. x = 4 sin t, y = 5 cos t, < t < 2ir 



16 sin 2 t i 25 cos 2 t 



16 



25 



1 



*- + 2- = 1 

16 T 25 1 





Copyright (c) 2006 Pearson Education 




154 Chapter 3 Differentiation 

71. x = 3t, y = 9t 2 , -oo < t < oo => y = x 2 





/ 


t 


y = * 2 / 




i<o\ 


^ 1 




f t>0 











i 





72. x = -Vt,y = t, t > => x 
or y = x 2 , x < 



,t>o 



x— Vy 



H i 1 1- 



f? ■ ■ X 



73. x = 2t - 5, y = 4t - 7, -oo < t < oo 
=>■ x + 5 = 2t =4> 2(x + 5) = 4t 

=>■ y = 2(x + 5) - 7 =>• y = 2x + 3 



74. x = 3 - 3t, y = 2t, < t < 1 =>■ § = t 

=>• x = 3-3(|) => 2x = 6-3y 
=> y = 2 - | x, < x < 3 





. 


. 








4 


/ 






/ 


= 5/2 


- 


j = 2x + 3 




J =7/4. 


/-l 
-1 

-2 

-3 

-4 




12 3 4 





y-2-(2x/3) 
1-1 



1 — I — I — I 1 — t-^ — I- 



75. x = t, y = \f\ -t 2 , — 1 < t < 
=> y = \f\ - x 2 



y = A/l-j: 2 



76. x = v/t+ 1, y = yfx, t > 

=4> y 2 = t => x = vV + 1, y > 

"-Vy^T . 



\1 

-\ — i — i — t-X 



y t>o 



" l ! ■ X 



77. x = sec 2 t - 1, y = tan t, - § < t < | 
=>• sec 2 1 — 1 = tan 2 1 => x = y 2 





y 








/ 


V 








3 






< < < ir/2 




2 










1 
f = 






* = / 


-2 


-1 

-1 

-2 


I 1 


2 


3 4 ' A 




-3 






-f S ,<0 



78. x = - sec t, y = tan t, - f < t < | 

=> sec 2 t - tan 2 1 = 1 =4> x 2 - y 2 = 1 



v 0<t<*/2 ..2 2 . ,' 
x -y .1 / 




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Section 3.5 The Chain Rule and Parametric Equations 155 



79. (a) x = a cos t, y = — a sin t, < t < 27r 

(b) x = a cos t, y = a sin t, < t < 2ir 

(c) x = a cos t, y = — a sin t, < t < 4ir 

(d) x = a cos t, y = a sin t, < t < 4n 



80. (a) x = a sin t, y = b cos t, | < t < ^ 



(b) x 

(c) x 

(d) x 



b sin t, < t < 2tt 



a cos t, y = 
a sin t, y = 
a cos t, y = b sin t, < t < 4ir 



b cos t, § < t < 9 f 



81. Using (—1, —3) we create the parametric equations x = — 1 + at and y = — 3 + bt, representing a line which goes 
through (—1, —3) at t = 0. We determine a and b so that the line goes through (4, 1) when t = 1. 

Since 4=— l+a=^>a=5. Since 1 = — 3 + b =^b = 4. Therefore, one possible parameterization is x = — 1 + 5t, 
y = -3 - 4t, < t < 1. 

82. Using (—1, 3) we create the parametric equations x = — 1 + at and y = 3 + bt, representing a line which goes through 
(— 1, 3) at t = 0. We determine a and b so that the line goes through (3, —2) when t = 1. Since 3 = — 1 + a =^ a = 4. 
Since —2 = 3 + b =4> b = —5. Therefore, one possible parameterization is x = — 1 + 4t, y = — 3 — 5t, < t < 1. 

83. The lower half of the parabola is given by x = y 2 + 1 for y < 0. Substituting t for y, we obtain one possible 
parameterization x = t 2 + l,y = t, t < 0. 

84. The vertex of the parabola is at (—1, —1), so the left half of the parabola is given by y = x 2 + 2x for x < —1. Substituting 
t for x, we obtain one possible parametrization: x = t, y = t 2 + 2t, t < —1. 



85. For simplicity, we assume that x and y are linear functions of t and that the point(x, y) starts at (2, 3) for t = and passes 
through (-1, -1) at t = 1. Then x = f(t), where f(0) = 2 and f(l) = -1. 

Since slope = % = z jE^ = -3, x = f(t) = -3t + 2 = 2 - 3t. Also, y = g(t), where g(0) = 3 and g(l) = -1. 
Since slope =% = =££ = -4. y = g(t) = -4t + 3 = 3 - 4t. 
One possible parameterization is: x = 2 — 3t, y = 3 — 4t, t > 0. 



86. For simplicity, we assume that x and y are linear functions oft and that the point(x, y) starts at (— 1, 2) for t = and 
passes through (0, 0) at t = 1. Then x = f(t), where f(0) = -1 and f(l) = 0. 

Since slope =% = ^j^r = 1 > x = f ( l ) = ll + (- 1 ) = -1 + 1 Also, y = g(t), where g(0) = 2 and g(l) = 0. 
Since slope = ^ = ^ = -2. y = g(t) = -2t + 2 = 2 - 2t. 
One possible parameterization is: x = — 1 + t, y = 2 — 2t, t > 0. 



87. t = \ =>■ x = 2 cos \ = \fl, y = 2 sin \ 



v^;| 



-2 sin t, % 

' at 



=>■ 


m =_ co t£ = 

dx w 4 


=>■ 


d 2 y dy'/dt esc 2 1 

dx 2 — dx/dt — -2 sin t 




If => x = cos \ = 



1 ; tangent line is y 

d 2 y I 
dx 2 



2 sin 3 1 



dy 
dx 

dM 

dx 2 



i,y= v^cosf = 
-\/3 ; tangent line is y — ( — , ) 



'2= -1 (x 
—y/2 



y/3 . dx 

2 ' dt 



2 cos t 

2 ) or y = - 



dy 
dx 



dy/dt _ 2 cos t 
dx/dt — 2 sin t 



COtt 



2y2;^=csc 2 t 



sint, ^ 

' dt 



V~3 



**t => | 



y3[x-(-i)]ory= V / 3x 



. d/ 
dt 



-y/3 sint 



V~3 



cry 
dx 2 



dx 
dt 



x ' dt 



i 



dy 
dx 



I = l-(x-I)or y = x+i;f 



4 L 



dy/dt 
dx/dt 

d 2 y 
* dx 2 



27t 

dy'/dt 
dx/dt 



dy 
dx 



1 ; tangent line is 



1 <-3/2 _v d^yl 
4 L ^ dx 2 



Copyright (c) 2006 Pearson Education 




156 Chapter 3 Differentiation 



90. t = 3 => x = - v / 3+T = -2, y = y / 3(3) = 3; 

_ _ 3y/t+T _ dy I _ -3y/3 + l 



(t+1) -l/2 ) | = | (3t) -l/2 



dy 



(l)w-^ 



/3t dx | (=3 



/3(3) 



dt 2 V-V "^ dx (_l)( t+ l)-l/2 

2; tangent line is y — 3 = — 2[x — (—2)] or y = — 2x — 1; 



d/ _ V^ [- | (t + l)- 1 / 2 ] + 3 Vm [§ (3t)-V 2 ] 3 . <Py 

dt 3t 2t,/3t\A+T dx2 



v2tV3t ,/t+l 



d 2 y| 

dx 2 



jh) 



3 

t73t 



91. t= -1 => x = 5,y = 1; 
y — 1 = 1 • (x — 5) or y = 



dx _ A* dy _ 4f 3 

dt — ^ L ' dt ~~ ^ l 



4; 



d/ 



2t 



dy dy/dt 4t^ t 2 

dx ~~ dx/dt — ^47 — 



d 2 y _ dy'/dt _ 2t 
dx 2 — dx/dt — 4t 



? dx| 

dfyl 
dx 2 . 



(— l) 2 = 1; tangent line is 



92. t 



dy 
dx 



3 

(!) 



4.i 



1 



1 



1 . dx _ 1 

2 ' dt ~~ l 



cos t, -i = sin t 

' dt 



v/3x 



-l 

(1 -cost) 2 



3 

d 2 y I 
dx 2 



1— (!) " (I) 

(1 — cos t)(cos t) — (sin t)(sin t) 



V3 ; tangent line is y — | = y3 ( x ~ f + 2 ) 



+ 2;1 



d/ _ 
dt 

-4 



(1-cost) 2 



d^v 

dx-' 



dy7dt 
dx/dt 



dy 
dx 



dy/dt 
dx/dt 



93. t 



dy 
dx 



COS 



0,y= 1 



sin 



9- dx 

z ' dt 



sin t 



dy 



cos t => 



dy 



cot f=0; tangent line is y = 2: 



d/ 



d 2 y 
dx 2 



dx 

CSC 2 t 



COtt 

dfyl 

" dx 2 



94. t 



dy 
dx 



X = 

sec 2 t 



2 sec 2 1 tan t 



sec 2 (-f) -1 = l,y 
cott => 



4/ 
1 _ 1 



2tant 



; tan 

dy 
dx . 



I) 



1 . dx 
l ' dt 



2 sec 2 1 tan t, $ = sec 2 t 



cot I 



l . 

2 



; tangent line is 



y-(-l)=-i(x-l)ory=-ix-!;f 



1 csc 2 f . d^y 

2 CSC I =? dx2 



2 sec 2 t tan t 



\ COt 3 t 



d 2 y I 
dx 2 



95. s = A cos (27tbt) => v = f t = -A sin (27rbt)(27rb) = -27rbA sin (27rbt). If we replace b with 2b to double the 
frequency, the velocity formula gives v = — 47rbA sin (47rbt) =>• doubling the frequency causes the velocity to 
double. Also v = -27rbA sin (27rbt) =>- a = ^ = -47r 2 b 2 A cos (27rbt). If we replace b with 2b in the 
acceleration formula, we get a = — 167r 2 b 2 A cos (47rbt) => doubling the frequency causes the acceleration to 



quadruple. Finally, a = — 47r 2 b 2 A cos (27rbt) =>• j 



da 
dt 



87r 3 b 3 A sin (27rbt). If we replace b with 2b in the jerk 



formula, we get j = 64-7r 3 b 3 A sin (4-7rbt) => doubling the frequency multiplies the jerk by a factor of 8. 



96. (a) y = 37 sin [|| (x - 101)] + 25 



y 



37 cos [fa (x - 101)] 



' 2tt n 
v 365 / 



747r COS \& (X • 



365 



.365 



101) 



The temperature is increasing the fastest when y' is as large as possible. The largest value of 

cos [M= (x — 101)] is 1 and occurs when ^ (x — 101) = => x = 101 => on day 101 of the year 

( ~ April 11), the temperature is increasing the fastest. 



(b) y'(lOl) 



365 



cos [|L ( ioi -101)] 



365 



cos (0) 



74^ 
365 



0.64 °F/day 



97. s = (1 + 4t) 1 / 2 => v 
v = 2(1 + 4t) _1/2 => 



ds _ 1 

dt 2 

- dv _ 

— dt " 



(1 + 4t)~ 1/2 (4) = 2(1 + 4t)- 1 / 2 => v(6) = 
-- - \ - 2(1 + 4t) _3/2 (4) = -4(1 + 4t)" 3 / 2 



2(1 + 4 • 6)" 1 / 2 = | m/sec; 



a(6)= -4(1+4-6) 



-3/2 



m/sec 



Copyright (c) 2006 Pearson Education 




We need to show a 



dv 
dt 



is constant: a 



Section 3.5 The Chain Rule and Parametric Equations 



% = %■ • f and ^ 

dt as dt as 



k 

2, A 



dv ds 
ds ' dt 



dv 
ds 



2\A 



■Ws 



k 2 



which is a constant. 



157 



99. v proportional to -k- => v = -7- for some constant k 

V s V s 



dv 
ds 



2s ; J2 



Thnc a — dv — dv . ds — 9X . v 

inus, a— , _ — • — _ — - v 



dt ds dt 



ds 



k k 

2 S 3/2 ' ^ 



^ (J 

2 \s : 



(-5) =4> acceleration is a constant times \ so a is inversely proportional to s 2 



100. Let f = f(x). Then, a 



dv dv dx 

dt dx dt 



dv 
dx 



f « = S (f ) * f « = l ( f ( x )) • f W = f '( x ) f ( x )> as required. 



101. T = 27r,/t => 5i = 2tt 

1 dL 



■f- . Therefore, & = 2£ .& = -£- -iL = ^# = 1 . 27rk A /t 

/gL du dL du ,/gL ^g 2 y g 



^ , as required. 



102. No. The chain rule says that when g is differentiable at and f is differentiable at g(0), then f o g is 
differentiable at 0. But the chain rule says nothing about what happens when g is not differentiable at so 
there is no contradiction. 

1 03 . The graph of y = (f o g)(x) has a horizontal tangent at x = 1 provided that (f o g)'( 1 ) = =4> f '(g( 1 ))g'( 1 ) = 

=> either f'(g(l)) = or g'(l) = (or both) =>■ either the graph of f has a horizontal tangent at u = g(l), or the 
graph of g has a horizontal tangent at x = 1 (or both). 

104. (fog)'(— 5)<0 =$■ f'(g(-5)) • g'(-5) < =^ f'(g(-5)) and g'(— 5) are both nonzero and have opposite signs. 
That is, either [f (g(-5)) > and g'(-5) < 0] or [f (g(-5)) < and g'(-5) > 0] . 



105. As h — y 0, the graph of y 



sin 2(x+h)— sin 2x 



approaches the graph of y = 2 cos 2x because 
(sin 2x) = 2 cos 2x. 



lim 

h->0 



sin 2(x+h)— sin 2x 



dx 




O.i i r cos f(x + h) 2 l— cos (x 2 ) 
, the graph of y = — l - ^ ^- L 



106. Ash 

approaches the graph of y = — 2x sin (x 2 ) because 

[cos (x 2 )] = — 2x sin (x 2 ). 



lim 

h->0 



cos [(x + h) 2 ] —cos (x 2 ) 
h 



_d_ 

dx 




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158 Chapter 3 Differentiation 



107. f = cos t and f = 2 cos 2t 



dy 
dx 



dy/dt 
dx/dt 



2 cos 2t 



2(2cos 2 t-l) . 

cos t ' 



theng=0 



2 (2 cos 2 1- 1) 
cos t 



y? 



=4> 2 cos 2 1 - 1 = => cos t = ±4= => t = f , ^ , ^ , ^ . In the 1st quadrant: t = | => x = sin | - ., 
y = sin 2 (|) = 1 =>• I -^-, 1 1 is the point where the tangent line is horizontal. At the origin: x = and y = 



and 



=>- sin t = => t = or t = 7r and sin 2t = =^ t = 0, 



3tt . 



the origin. Tangents at origin: 
& - 3 cos 3t 



2x and 



thus t = and t = 7r give the tangent lines at 



-2x 



108. f = 2 cos 2t and d( 



dy dy/dt 3 cos 3t 3(cos 2t cos t — sin 2t sin t) 

dx — dx/dt — 2 cos 2t ~~ 2(2cos 2 t-l) 

_ 3 [(2 cos 2 t- 1) (cos t)- 2 sin t cost sin t] _ (3 cos t) (2 cos 2 t - 1 - 2 sin 2 1) _ (3 cos t) (4 cos 2 t - 3) 
~ 2(2cos 2 t-l) ~~ 2(2cos 2 t-l) ~~ 2(2cos 2 t-l) 

dy =Q (3 cos t) (4 cos 2 1 - 3) =Q 3 C os t = or 4 cos 2 1 - 3 = 0: 3cost = => t=f , ^fand 

dx 2(2 cos 2 t— 1) 2 ' 2 



; then 



4 cos 2 1 - 3 = =>• cos t = ± 



>/5 



t 



7T 57T ITT 117T 



In the 1 st quadrant: t 



sin 



2(1) 



x/3 



and y = sin 3 (|) = 1 =>• ( ^-, 1 ) is the point where the graph has a horizontal tangent. At the origin: x = 



and y = =4> sin 2t = and sin 3t = => t = 0, | , n, y and t = 0, 
the tangent lines at the origin. Tangents at the origin: j- 



7T 2tt 4_7r 5n 

3 ' 3 ' 7r ' 3 ' 3 



2 cos 



§ x, and 5Z 

2 ' dx 



t = and t = 7r give 

dy 



3 cos (3tt) 
2 cos (27r) 



3 X 
2 X 



109. From the power rule, with y = x 1 / 4 , we get -=- 



dy 



j x 3//4 . From the chain rule, 



dx 



*'*<^» 



1 1 

' 2 X A 



I x -3/4_ 



2 Vv^ 



in agreement. 



1 10. From the power rule, with y = x 3 / 4 , we get -r- = § x 1//4 . From the chain rule, y 



2 V x V x 

3 \A _ 3 „-l/4 



dy 
dx 



^V^vV* 



£K^) 

in agreement. 



2,/xVx 



Xa/X 



bx 1 ' ( x ' 2^ + 



2,/x,/x 



^) 



3y^ 
4,/xT/x 



111. (a) 



dg/dt 



IP- 



-ir/2 
O— 



w/2 

-1 O- 



TT 

-o 



(b) f t = 1.27324 sin 2t + 0.42444 sin 6t + 0.2546 sin lOt + 0.18186 sin 14t 

(c) The curve of y = 4j approximates y = -J 



the best when t is not — w, 



dt 
0, f , nor 7r. 



df/dt 




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Section 3.5 The Chain Rule and Parametric Equations 159 



112. (a) 



dg/dt 



-IT -IT/2 



w/2 Tf 



(b) f = 2.5464 cos (2t) + 2.5464 cos (6t) + 2.5465 cos (lOt) + 2.54646 cos (14t) + 2.54646 cos (18t) 

(c) 

dh/dt 




_t 



111-116. Example CAS commands: 
Maple : 

f := t -> 0.78540 - 0.63662*cos(2*t) - 0.07074*cos(6*t) 
- 0.02546*cos(10*t) - 0.01299*cos(14*t); 

g := t -> piecewise( t<-Pi/2, t+Pi, t<0, -t, t<Pi/2, t, Pi-t ); 

plot( [f(t),g(t)], t=-Pi..Pi ); 

Df := D(f); 

Dg := D(g); 

plot( [Df(t),Dg(t)], t=-Pi..Pi ); 
Mathematica : (functions, domains, and value for tO may change): 
To see the relationship between f[t] and f [t] in 111 and h[t] in 112 

Clear[t, f] 

f[t_] = 0.78540 - 0.63662 Cos[2t] - 0.07074 Cos[6t] - 0.02546 Cos[10t] - 0.01299 Cos[14t] 

f[t] 

Plot[{f[t],f[t]},{t,-7r,7r}] 
For the parametric equations in 1 13 - 1 16, do the following. Do NOT use the colon when defining tanline. 

Clear[x, y, t] 

tO = p/4; 

x[t_]:=l-Cos[t] 

y[t_]:=l+Sin[t] 

pl=ParametricPlot[{x[t], y[t]},{t, -7T, 7r}] 

yp[t_]:=y'[t]/x'[t] 
ypp[t_]:=yp'[t]/x'[t] 

yp[t0]//N 

ypp[t0]//N 

tanline [x_]=y[t0] + yp[t0] (x - x[t0]) 

p2=Plot[tanline[x], {x, 0, 1}] 

Show[pl, p2] 



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160 Chapter 3 Differentiation 
3.6 IMPLICIT DIFFERENTIATION 



1 v _ x 9/4 =, dy _ 9 5/4 

i . y — x ^ dx _ 4 x 



2. y = X - 3 / 5 =>. g = _| x -8/5 



3. y = ^2x" = (2x)V3 =► g = § (2x)- 2 / 3 ■ 2 = ^ 



4. y = ^5x" = (Sx) 1 ^ => g = i (5x )-3/4 .5 = 1^ 



5. y = 7 v / x + 6 = 7(x + 6) 1 / 2 => | = | (x + 6)- 1 / 2 - 7 



2-v/x + 6 



6. y = -2x/x^T = -2(x - l) 1 / 2 => g = -l(x - 1)" 



1/2 



7. y = (2x + 5)- 1 /' 2 => g = - I (2x + 5)- 3 / 2 - 2 = -(2x + 5)- 3 / 2 



y = (1 - 6x) 2 / 3 =>■ g = | (1 - 6x)- 1 / 3 (-6) = -4(1 - 6X)- 1 / 3 



9. y = x (x 2 + 1) 1/2 => y' = x • i(x 2 + l)~ 1/2 (2x) + (x 2 + 1) 1/2 • 1 = (x 2 + l)" 1/2 (x 2 + x 2 + 1) = ^=±± 



10. y = x (x 2 + 1)- 1/2 => y' = x • (-i)(x 2 + l)- 3/2 (2x) + (x 2 + 1)~ 1/2 ■ 1 = (x 2 + l)- 3/2 (-x 2 + x 2 + 1) -- ■ , . , 



l 

( X 2+l) 3 



11 s - , 7 /t2 - t 2 / 7 ^> ds - 2 t -5/7 
11.;,— V L — i ^ dt — 7 l 



12. r 



-3 = fl-3/4 =, dr 



3 0-7/4 



13. y = sin ((2t + 5r 2 / 3 ) => f = cos ((2t + 5r 2 / 3 ) • (- f ) (2t + 5ir 5 / 3 - 2 = - f (2t + 5r 5 / 3 cos ((2t + 5)~ 2 



-2/3\ 



14. z = cos ((1 - 6t) 2 / 3 ) =>- | = -sin ((1 - 6t) 2 / 3 ) - | (1 - 61)- 1 / 3 (-6) = 4(1 - 61)- 1 / 3 sin ((1 - 6t) : 



ft) 



15. f(x) = J 1 



,1/2 



-1/2 



(l-x 1 / 2 ) 1 ^ => f'(x)=I(l-x 1 / 2 ) _1// (-ix- 1 / 2 ) 



'(V 1 -^)^ V x ('-v^) 



16. g(x) = 2 (2x-V 2 + 1)~ 1/3 =► g'(x) = - | (2X-V2 + l)~ 4/3 - (-l)x- 3 / 2 = I (2X- 1 / 2 + 1)~ 4/ V 3 / 2 

17. h(0) = j/l + cos (26») = (1 + cos 20) 1 / 3 =>• h'(0) = I (1 + cos 26») 2 / 3 • (-sin 20) ■ 2 = - § (sin 25) (1 + cos 26>r 2/3 



18. k(0) = (sin (0 + 5)) 5/4 =>• k'(0) = § (sin (5 + 5)) 1/4 • cos(0 + 5) = I cos(0 + 5)(sin(0 + 5)) 1 



/4 



19. x 2 y + xy 2 = 6: 



Stepl: (x 2 g+y-2x) + (x-2yg+y 2 -l)=0 

Step 2: X 2g + 2x y g = -2xy-y 2 
Step 3: g (x 2 + 2xy) = -2xy - y 2 



dx 

Step 4: g = ^£ 



20. x 3 + y 3 = 18xy =>- 3x 2 + 3y 2 g = 18y + 18x g =>• 



(3y 2 - 18x) g = 18y - 3x 2 =► g = Jf=| 



21. 2xy + y 2 = x + y: 

Stepl: (2xg+2y)+2yg = l 



dy 
dx 



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Section 3.6 Implicit Differentiation 161 



Step 2: 2xg+2yg-g = l-2y 
Step 3: g(2x + 2y-l) = l-2y 



Step 4: g 



dx 

dy _ 

dx 2x + 2y — 1 



1 - 2y 



22. x 3 - xy + y 3 = 1 =* 3x 2 - y - x g + 3y 2 g = => (3y 2 - x) g = y - 3x 2 =^| = ^ 



23. x 2 (x-y) 2 



Step 1: x 2 [2(x - y) (l - g)] + (x - y) 2 (2x) = 2x - 2y g 

Step 2: -2x 2 (x - y) g + 2y g = 2x - 2x 2 (x - y) - 2x(x - y) 2 

Step 3: g [-2x 2 (x - y) + 2y] = 2x [1 - x(x - y) - (x - y) 2 ] 

Stpn 4- ^ — 2x [ 1 -"( x -y)-( x -y) 2 ] _ x[l-x(x-y)-(x-y) 2 ] _ x (1 - x 2 + xy - x 2 + 2xy - y 2 ) 

OLCpt. dx — -2x 2 (x-y) + 2y ~~ v -x2r Y _ v i T 2v_ ¥ 8_ 



y-x 2 (x-y) 



x^y — x J + y 



x — 2x 3 + 3x 2 y — xy 2 
x 2 y — x 3 +y 



24. 



(3xy + If = 6y => 2(3xy + 7) - (3x g + 3y) = 6 g => 2(3xy + 7)(3x) g - 6 g = -6y(3xy + 7) 



g [6x(3xy + 7) - 6] = -6y(3xy + 7) 



dy 
dx 



y(3xy + 7) 



3xy 2 + 7y 



x(3xy + 7) - 1 1 - 3x 2 y - 7x 



25. y 



2 _ x-1 

x+1 



26. x 2 = S=Z 

x + y 



9 dy _ (x+l)-(x-l) _ 2 . dy _ 1 

*J dx (x+l) 2 (x+1) 2 ~^ dx y(x+l) 2 

x 3 + x 2 y = x - y => 3x 2 + 2xy + x 2 y' = 1 - y' =4> (x 2 + 1) y' = 1 - 3x 2 - 2xy =>- y' =■ '■■-"-'■■-■-> 



x 2 +l 



27. x = tany => l = (sec 2 y)g => g = O^ = cos 2 y 



28. xy = cot (xy) =>• xg + y = -csc 2 (xy)(xg + yj => xg + xcsc 2 (xy)g = -y csc 2 (xy) - y 
■ [x + xcsc 2 (xy)] = — y [csc 2 (xy) + l] 



dy I 
dx I 



dy _ -y [csc 2 (xy) + l] 
dx x[l+ csc 2 (xy)l 



29. x + tan (xy) = =S> 1 + [sec 2 (xy)] (y + x g) = => x sec 2 (xy) g = - 1 - y sec 2 (xy) 



-l 



-cos 2 (xy) y —cos 2 (xy) — y 



x sec 2 (xy) x 



dy 
dx 



- 1 — y sec 2 (xy) 
x sec 2 (xy) 





a t sin y — *.y —r it v , -<-' a J ' dx ~ J ~ dx v ' ' dx ~ ■> 


dx cos y — X 


31. 


ysin( y )=l-xy =» y [cos (l) - (-1) £ - g] + S in (l) - g = -x g ■ 


-y => 




dy 1" 1 en- (A 1 Hn f 1 ] 1 xl v=t dy " y 


-y 2 




dx[ y C0 HyJ ' Jm \y) ' X J >^dx _i cos (i) +sin (i) +x ysin 


(i)-cos(l)+xy 


32. 


y 2 cos(i)=2x + 2y=.y 2 [-sin(i).(-l)i.g]+cos(l).2yg = 
dy [-inf lN | \^vca-(A ">1 " -- dy 2 


2 + 2g => 




dx^ m ^yj 1 -> C0 ^yJ -J -- dx sin (l) +2ycos (l)_ 2 




33. 


fll/2 i r l/2 _ 1 _v 1 fl-1/2 , 1 .-1/2 dr _ n -^ dr [ 1 1 _ -1 . dr _ 


2 \/r _ V^ 
2^9 v^ 



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162 Chapter 3 Differentiation 

34. r - 2^9 = § 2 / 3 + | 3 / 4 => | - 



-1/2 = -l/3 + -l/4 ^ dr = 0-1/2 + 0-1/3 + 0-1/4 



35. sin(r0) = i => [cos (r0)] (r + %) = =^ § [0 cos (r0)] 
cos (r0) / 



-r cos (r0) 



dr _ -r cos (rfl) 
dfl ~~ 0cos(r0) 



36. cos r + cot 9 = rd =4> (-sin r) %. - esc 2 6 = r+9 % => ^ [-sin r - 0] = r + esc 2 =4> ^ = -- 



37. x 2 + y 2 = l => 2x + 2yy' = =* 2yy' = -2x => g = y' = - f ; now to find § , £ (y') = £ (- |) 



,// _ y(-l)+xy' 



-y+x - i 



-y — - since y 



d 2 y // — y 2 — x 2 — y 2 — (1— y 2 ) — 1 



dx : 



i = y 



38. x 2 / 3 + y 2 / 3 = 1 =>• | x- 1 / 3 + | y- 1 / 3 g = => g [| y" 1 / 3 ] = - f x-V3 => y ' 

;l/3. (_ 1 y -2/3) y/ + yl/8 (1 x -2/3j ^ xV3.(-ly- 2 /3) f_ !^ J + yl/3 ( | x" 2 / 3 ) 



dy _ _ x^ 
dx y- 



,1/3 



Differentiating again, y" 



X 2 / 3 



^73 



d 2 ! _ 1 -2/3 v -l/3 i 1 v l/3„-4/3 _ ±i 



dx 2 3 



x -Z/O y -l /a + iyi/^X 



3x'l/3 ^ 3yl/3 x 2/3 



39. y 2 = x 2 + 2x => 2yy' = 2x + 2 =>• 

_, dfy _ „ _ y 2 - (x+ l) 2 
^ dx 2 — y — y3 



y = 2x + 2 = x+i ± 

•> 2y y ' J 



„ _ y-(x+l)y' 

y ; 



y-(x+Dp±i 



40. y 2 - 2x = 1 - 2y =4> 2y • y' - 2 = -2y' => y'(2y + 2) = 2 => y' = -L- = (y + l)" 1 ; then y" 

-i 
(y- 1)3 



-(y + i)" 2 • y' 



-(y+D-^y+i)- 1 => S=y"~ - 



41. 2,/y" = x - y => y-!/2 y ' = i _ y ' =j>. y ( y -i/a + l) = l => g = y 



^ 



y-1/2+1 - ^+1 . ""^ 

differentiate the equation y' (y~ 1/2 + l) = 1 again to find y": y' (- \ y" 3 / 2 y') + (y~ 1/2 + l) y" = 



(y-l/2 + 1 )y = I [y f y -3/2 ^ g = y, 



» l^Tj y " 3/2 



42. xy + y 2 = 1 => xy' + y + 2yy' = =4> xy' + 2yy' = -y =>• y'(x + 2y) = -y =>• y' - — --- • — 



(y-l/2+l) 2y3/2(y-l/ 2 +l) 3 2(l + y^) 3 

~y • : L - V 

(x+2y) ' dx 2 J 



-(x + 2 y )y' + y(l+2y') _ ~(* + 2y) [u+lyi] + y [' + 2 ((I+lyl)] 
(x + 2y) 2 (x + 2y) 2 

2y(x + 2y) - 2y 2 _ 2y 2 + 2xy _ 2y(x + y) 
(x + 2y) 3 " (x + 2y) 3 ~~ (x + 2y) 3 



(^>[y(x + 2y) + y(x + 2y) - 2y 2 ] 



(x + 2y) 2 



43. x 3 + y 3 = 16 => 3x 2 + 3y 2 y' = =4> 3y 2 y' = -3x 2 => y' = - K ; we differentiate y 2 y' = -x 2 to find y": 

9 -2x-2yf-4) 2 -2x-?4 

y 2 y" + y' [2y - y'] = -2x =* y 2 y" = -2x - 2y [y'] 2 => y" = J^- = — ^ 



-2xy 3 - 2x 4 d/V I 

yS ^ dx 2 



-32-32 
32 



44. xy + y 2 = l => xy' + y + 2yy' = => y'(x + 2y) = -y => y' = ^ => y = (x±Mlzgk^)(i±jyO 
since y| = - 1 we obtain f\ = LMIizlIM = _ l 



45. y 2 + x 2 = y 4 - 2x at (-2, 1) and (-2, -1) =» 2y g + 2x = 4y 3 g - 2 =» 2y g - 4y 



4x _ /u,3 dy 



-2-2x 



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Section 3.6 Implicit Differentiation 163 



f, (2y - 4y3) 



-2-2x 



dy 
dx 



x+1 
2y 3 -y 



dy 
dx 



-land^ 

dx 



46. (x 2 + y 2 ) 2 = (x-y) 2 at(L0) and (1,-1) => 2 (x 2 + y 2 ) (2x + 2y |) = 2(x - y) (l - &) 



dx [2y(x 2 + y 2 )-,-(x-y)] = -2x(x 2 + y 2 ) + (x-y)^ g = ^g^^ ' - 



Jy 



dy 



and 



dy 



47. x 2 + xy - y 2 = 1 =>• 2x + y + xy' - 2yy' = =>■ (x - 2y)y' = -2x - y => y 



/ 2x + y . 

2y — x ' 



(a) the slope of the tangent line m = y' | „ 3 = | =>• the tangent line is y — 3 = | (x — 2) 

(b) the normal line is y — 3 = — ^ (x — 2) => y = — ^ x + y 



48. x 2 + y 2 = 25 =>■ 2x + 2yy' = =>• y' = - 
(a) the slope of the tangent line m = y' | _ _ 4) 



y ' 



the tangent line isy + 4= | (x — 3) 



25 



(b) the normal line is y + 4 = — ^ (x — 3) =>• y 



x / y / 



=> 2xy 2 + 2x 2 yy' = => x 2 yy' = -xy 2 

(a) the slope of the tangent line m = y'| (1 3) = — | 

=> y = 3x + 6 

(b) the normal line is y — 3 = — | (x + 1) => y = 



3 =>■ the tangent line is y — 3 = 3(x + 1) 



50. y 2 - 2x - 4y - 1 = => 2yy' - 2 - 4y' = => 2(y - 2)y' = 2 => y' = -^ ; 

(a) the slope of the tangent line m = y'| 2 ,. = — 1 =>• the tangent line is y — 1 = — l(x + 2) 

(b) the normal line is y — 1 = l(x + 2) =>• y = x + 3 



-x- 1 



51. 6x 2 + 3xy + 2y 2 + 17y - 6 = => 12x + 3y + 3xy' + 4yy' + 17y' = => y'(3x + 4y + 17) = -12x - 3y 



/ _ -12x-3y . 

y 3x + 4y+17 ' 



(a) the slope of the tangent line m = y'| , „ = , ? ~ y 7 = f =>• the tangent line is y — = I (x + 1) 

(b) the normal line is y — = — ? (x + 1) =>■ y=-^x- ^ 



52. x 2 - v^xy + 2y 2 = 5 =* 2x - v^xy' - v^y + 4yy' = => y' (4y - \/3x) = V^y - 2x =* y' - - - - - . 



V / 3y-2x 



A —2 I 

(a) the slope of the tangent line m = y' I / R _\ = v ~ y ^ x 

(b) the normal line is x = \J 3 



=>• the tangent line is y = 2 



-2y 



53. 2xy + 7r sin y = 27r =>■ 2xy' + 2y + 7r(cos y)y' = => y'(2x + 7r cos y) = — 2y =4> y' — ,, 
(a) the slope of the tangent line m = y'L „ = , ~ y = — | =>■ the tangent line is 



I (i,f) 2x + 7r cos y 



(M) 



- | (x — 1) =4> y=-fx + 7r 



(b) the normal line is y — | = - (x — 1) => y = -x — - + f 



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164 Chapter 3 Differentiation 



54. x sin 2y = y cos 2x =>• x(cos 2y)2y' + sin 2y = — 2y sin 2x + y' cos 2x =>■ y'(2x cos 2y — cos 2x) 



= -sin 2y - 2y sin 2x^y'= -^ 2y + 2 y S m 2x 

J J J cos 2x — 2x cos 2y ' 

(a) the slope of the tangent line m = y' I ,, ,, = sin ^y + 2y sin 2x I 

v / r & ^ I (j,f J cos 2x — 2x cos 2y /J, jr\ 

y-| = 2(x-f) =* y = 2x 



^ = 2 =>- the tangent line is 



(b) the normal line is y 



Hx 



55. y = 2 sin (7rx — y) => y' = 2 [cos (7rx — y)] • (n — y') => y'[l + 2 cos (7rx — y)] = 2ir cos (irx — y) 



/ 2tt cos (ttx — y) . 

J 1+2 cos (7rx — y) ' 



(a) the slope of the tangent line m = y' | (1 0) = l + 2 "os I*x - ) = 27r => the tangent line is 



y - = 2tt(x - 1) => y = 2ttx - 2n 



(b) the normal line is y — = — ^- (x — 1) =>- y = — j- + y- 



56. x 2 cos 2 y — sin y = => x 2 (2 cos y)(— sin y)y' + 2x cos 2 y — y' cos y = =4> y' [— 2x 2 cos y sin y — cos y] 
= -2x cos 2 y4y'= 92 2xcos2y 4 _ ; 

J J 2x A cos y sin y 4- cos y 

(a) the slope of the tangent line m = y' | ~ Nk " s N 



I (0,7r) 2x 2 cos y sin y + cos y 



=> the tangent line is y = -k 



(b) the normal line is x = 



57. Solving x 2 + xy + y 2 = 7 and y = =^ x 2 = 7 =$■ x= ± \/l => ( — \J 7, j and ( a/7, j are the points where the 
curve crosses the x-axis. Now x 2 + xy + y 2 = 7 =S> 2x + y + xy' + 2yy' = =4> (x + 2y)y' = — 2x — y 



y' = - 2^+1 ^ m = _ 2x + y ^ ^ 



x + 2y 



x + 2y 



slope at I — v 7, ) is m = %~ = —2 and the slope at I y 7, ) is 



-j=- = —2. Since the slope is —2 in each case, the corresponding tangents must be parallel. 



58. x 2 +xy + y 2 = 7 => 2x + y + x g + 2y g = =* (x + 2y) g = -2x - y =► g = ^ and | = ^ . 
(a) Solving ^ = =^ — 2x — y = =4> y = — 2x and substitution into the original equation gives 

x 2 + x(— 2x) + (— 2x) 2 = 7 => 3x 2 = 7 => x= ±a/I and y = T 2-i/ 1 when the tangents are parallel to the 



x-axis. 



■ ■ , - = 7 =^ — = 1 

21 ' \ 21 ' =? 4 ' 



(b) Solving f^=0 =^ x + 2y = =4> y = - § and substitution gives x 2 + x (- §) + ' x ^" - n 
=>• x = ± 2 J | and y = T yf when the tangents are parallel to the y-axis. 



59. y 4 = y 2 - x 2 =^> 4y 3 y' = 2yy' - 2x => 2 (2y 3 - y) y' = -2x => y' 



y-2y 3 



the slope of the tangent line at 



(4.$)* 



^ 



y-2y3 \(££\ ~ j/fTibZI - i_3 - 2 _ 3 

1 I 4 i 1 I 2 S "* 4 



1 ; the slope of the tangent line at (4,5) 



£. 



is 



2\/3 



V^ 



60. y 2 (2-x) = x 3 => 2yy'(2-x) + y 2 (-l) = 3x 2 => y' - f±H- • 



m 



y 2 -3x 2 I 
2y(2-x) 



y - 1 = - \ (X - 1) =4> y = -i x -, ., 



, s , . , the slope of the tangent line is 
2 => the tangent line is y — 1 = 2(x — 1) =4> y = 2x— 1; the normal line is 

1 „ 1 3 



61. y 4 - 4y 2 = x 4 - 9x 2 => 4y 3 y' - 8yy' = 4x 3 - 18x =>• y' (4y 3 - 8y) = 4x 3 - 18x =4> y 



/ _ 4x 3 - 18x _ 2x 3 - 9x 
4y 3 - 8y 2y 3 - 4y 



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Section 3.6 Implicit Differentiation 165 



x (2x 2 - 9) 
y(2y 2 -4) 



m; (—3,2): m 



(-3)(18-9) 
2(8-4) 



-f ;(-3,-2): m: 



27 . 



(3,2): m: 



27 . 



(3,-2): m=- 



27 



62. x 3 + y 3 - 9xy = =4> 3x 2 + 3y 2 y' - 9xy' - 9y = => y' (3y 2 - 9x) = 9y - 3x 2 =>• y' 

( a ) y'1,4.2, = I and y' 1 ,2,4) = ?; 



9y - 3x 2 
3y 2 - 9x 



3y-x 2 
y 2 — 3x 



(b) y' = => pEg = => 3y - x 2 = => y = f => x 3 + (f ) - 9x (f ) = => x 6 - 54x 3 = 

=*• x 3 (x 3 - 54) = =4> x = or x = V54 = 3 3 y / 2 => there is a horizontal tangent at x = 3 V 2 . To find the 
corresponding y-value, we will use part (c). 



(c) 







y 2 — 3x 
3y - x 2 



=> y 2 -3x = => y = ± a/3x ; y = v^x =>• x 3 



9xV3x = 

:? 



dy 

=> x 3 - 6a/3 x 3 / 2 = => x 3 / 2 (x 3 / 2 - 6\/3) = => x 3 / 2 = or x 3 / 2 = 6^ => x = or x = VT08 = 3 \A • 

Since the equation x 3 + y 3 — 9xy = is symmetric in x and y, the graph is symmetric about the line y = x. 
That is, if (a, b) is a point on the folium, then so is (b, a). Moreover, if y' | . b = m, then j/| . a) = ^ . 
Thus, if the folium has a horizontal tangent at (a, b), it has a vertical tangent at (b, a) so one might expect 
that with a horizontal tangent at x = y 54 and a vertical tangent at x = 3 y 4, the points of tangency are 



I v 54, 3 v 4 ) and I 3 y 4, y 54 ) , respectively. One can check that these points do satisfy the equation 

x 3 + y 3 - 9xy = 0. 



63. x 2 - 2tx + 2t 2 



2y 3 - 3t 2 



2x 



.,2 dy 



dx 



2x-2t 



6y 2 f t ~ 6t = 



dx 
dt 


+ 4t = 





=> (2x- 


2t)| = 


= 2x 


-4t 


=> 


dx 
dt " 


2x- 

" 2x- 


-4t 
-2t 


_ x-2t 

— x-t 


; 


dy 
dt 


6t 
~~ 6y 2 " 


_t 

r 


- ; *us g 


dy/dt 
~~ dx/dt 


_ (?) . 


" y 2 (> 


-t) 
,-2t) 


;t = 


2 






. + 4 = 


=^ 


(x - 2) 2 


= => 


X = 


2;t 


= 2 


=> 


2y 3 


- 


3(2) 2 = 


4 



2y 3 = 16 



y = 2; therefore -%■ 

J ' dx 



64. x = ^5 - v/i => I = I (5 - Vt) 

=► (t-i)i 



-1/2 



|t- i/2 ) 



2(2- 


-2) 


(2) 2 (2- 


-2(2)) 




1 



tv/t^5-,/t 



;y(t-l)=v/t => y +(t_ 1)| = 1^/8 



2\A 



dy _ yT y _ l-2y\A . , dy 
dt ~ (t-1) - 2ty/l-2y/i ' dx 



l-2s\._ 
2ly/t — 2\/t 

-1 

4\A l/5 " \A 



l-2yy/t 
2vA(t-l) 



4Vt\/5-Vt 



2(1-2x^)^5-^ 



therefore, 



l-t 



dy 



5- y r A=y r l; t = 4 => y(3) = ^ = 2 



i(l-2(2) V / 4)v / 5-V / 4 



1-4 



14 
3 



65. x + 2x 3 / 2 = t 2 + t 



dx 



dt + 3xV2 I = 2t + l =* (! + 3xl/2 ) I = 2t + ! 



dx _ 2t+l 



dT- TT 3^;y^+ T + 2t ^y = 4 

V / FTT + y(i)(t+l)-V2 + 2yjF+2t(iy-V2)| = 0^ | yTTT+ ^ + 2^+ (-^) £ = 

dy _ (yJTT- 2 A> -yyy-4yy^TT 



t+1 



v/y/ dt 2Vt+ 



fc- 2 v^ 



thus 



-y^/y-*y\/tTJ_ 
dy _ dy/dt _ V2^(i+l) + 2t s /FTT 
dx dx/dt / 2t + 1 \ 

^1+3x1/2; 



dt " (7F+T+ -Sj) " 2 v /^(t+l) + 2t^Tl 

; t = => x + 2x 3 / 2 = =^ x (l + 2x J / 2 ) =0 => x = 0;t = 



yy / Q+ 1 + 2(0)^ = 4 => y = 4; therefore 



-4^-4(4)^0 + 1 
2^(0+11 + 2(0)^0+7 

2(0)+ 1 ~\ 
l+3(0)V2i 



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166 Chapter 3 Differentiation 



66. x sin t + 2x = t 
t sin t — 2t = y 



dx 



sin t + x cos t 



9 dx 
z dt 



1 



sin t + t cos t 



(sint + 2)f = 
— • thus — — sin [ + ' cos ' ~ 2 • 



1 — X cos t 



dx _ 

dt sin t+2 



dt 



£ ; therefore ■/■ 

2 ' dx 



sin 7T + 7T COS 7T — 2 



(h)costt 



dx 



-4--S 

2 + 7T 



1 — X COS t 

t = 7T =>- X Sin 7T + 2X = 7T 



sin t + 2 

-4 



67. (a) if f(x) = | x 2 / 3 - 3, then f'(x) = x" 1 / 3 and f"(x) 

(b) if f(x) = ^ x 5 / 3 - 7, then f'(x) = | x 2 / 3 and f"(x) = x" 1 / 3 is true 



i x 4 / 3 so the claim f"(x) = x 1|/3 is false 



(c) f"(x) = \-^ 3 => f'"(x) = - i x~ 4 / 3 is true 

(d) if f'(x) = | x 2 / 3 + 6, then f"(x) = x" 1 / 3 is true 



68. 2x 2 + 3y 2 
also, y 2 = 



> 4x + 6yy' = 
2yy' = 3x 2 => \ 



3x 2 

2y 



2x 

3y 

y' 



(1,1) 


= - 


2x 

3y| 


3x 2 




3 


2y 


CM) 


2 



(1,1) 
and y' 



| andy'| (U _ n 



2x 

3y 



(i-i) 



3x^1 

2y 



2 . 

— 3 ' 
l-l) 

Therefore 



i (i,-D 

the tangents to the curves are perpendicular at (1, 1) and (1,-1) (i.e., the curves are orthogonal at these two 
points of intersection). 



69. x 2 + 2xy - 3y 2 = 
tangent line m = y" 



y'(2x - 6y) = -2x - 2y 



x + y 



2x + 2xy' + 2y - 6yy' = - , ._.. ,., , _. _, - -, - 3y _ x 

- — - ' — 1 => the equation of the normal line at (1,1) is y — 1 = — l(x — 1) 



the slope of the 



CM) 3y-*lc M) 

=> y = —x + 2. To find where the normal line intersects the curve we substitute into its equation: 
x 2 + 2x(2 - x) - 3(2 - x) 2 = => x 2 + 4x - 2x 2 - 3 (4 - 4x + x 2 ) = => -4x 2 + 16x - 12 = 
=^> x 2 — 4x + 3 = => (x — 3)(x — 1) = => x = 3 and y = — x + 2 = — 1. Therefore, the normal to the curve 
at (1, 1) intersects the curve at the point (3, —1). Note that it also intersects the curve at (1, 1). 



70. xy + 2x-y = =>• x^| + y + 2-^ = =>■ % = f±f ; the slope of the line 2x + y = is -2. In order to be 



parallel, the normal lines must also have slope of —2. Since a normal is perpendicular to a tangent, the slope of 
the tangent is i . Therefore, y^ = | =>• 2y + 4 = 1 — x =>• x = — 3 — 2y. Substituting in the original equation, 
y(-3 - 2y) + 2(-3 - 2y) - y = => y 2 + 4y + 3 = => y = -3ory = -l. If y = -3, then x = 3 and 
y + 3 = -2(x-3) =4> y = -2x + 3. If y = -1, then x = -1 and y + 1 = -2(x + 1) =4> y = -2x - 3. 



yi-o 



71. y=x =>• gj = ;r ■ If a normal is drawn from (a, 0) to (xi,yi) on the curve its slope satisfies ^— - = — 2yi 
=> yi = — 2yi(xi — a) or a = Xj + \ . Since xi > on the curve, we must have that a > \ . By symmetry, the 
two points on the parabola are (xj , a/x7) and (xi , — a/x+) ■ For the normal to be perpendicular, 



'i _a / V a_x i/ 



-1 => 



Therefore, 



? , ± j) and a 



(a- Xl )2 

_ 3 
~ 4 • 



1 =>- Xi = (a — Xi) 2 => Xi 



*1 



i _ Xl ) => X! = \ andyi 



72. Ex. 6b.) y = x 1//2 has no derivative at x = because the slope of the graph becomes vertical at x = 0. 

Ex. 7a.) y = (1 — x 2 ) has a derivative only on (— 1, 1) because the function is defined only on [— 1, 1] and 
the slope of the tangent becomes vertical at both x = — 1 and x = 1 . 



73. xy 3 + x 2 y = 6 

_ y 3 + : 

~~ 3xy 2 + x 2 



*(3y 2 £) 



.,3 jl v2 dy + 2xy 



y> -j- x 1 







(3xy 2 



2xy 



dx 



'—^- . a i SO; xy 3 + x 2 y = 6 ^ x ( 3y 2) + y 3 I + x 2 



y (2x I) = => I (y 3 + 2xy) 



— y 3 — 2xy 
3xy 2 + x 2 

-3xy 2 - 



dx 
dy 



3xy 2 + x 2 
y 3 + 2xy 



thus p- appears to equal -^ . The two different treatments view the graphs as functions 



dy 



Copyright (c) 2006 Pearson Education 




Section 3.6 Implicit Differentiation 167 

symmetric across the line y = x, so their slopes are reciprocals of one another at the corresponding points 
(a, b) and (b, a). 



74. x 3 + y 2 = sin 2 y => 3x 2 + 2y ^ = (2 sin y)(cos y) gf => ^ (2y - 2 sin y cos y) = -3x 2 => 



3x 2 



—w- ; also, x 3 + y 2 = sin 2 y 



3x 2 jjs + 2y = 2 sin y cos y 



^ — ■ ?r~ , (ALJKJ. j\. ~r y — axil v — f ~»./v ~r~ ~t~ ^1 — ^- am v ws V — f j~~ — t — o ■> mun j 

2 sin y cos y — 2y ' * J J dy J J J dy 5x z ' dy 

appears to equal A . The two different treatments view the graphs as functions symmetric across the line 

d7 

y = x so their slopes are reciprocals of one another at the corresponding points (a, b) and (b, a). 

(b) 



dy _ -3x 2 

dx 2y — 2 sin y cos y 

; thus 



75. x 4 + 4y 2 = 1: 
(a) y 



2 _ 1-x 1 
— 4 



±W\-x i 



dx 



±i(i-xT 1/2 (-4x 3 ) 



±X' J 



(l-x^) 1 '' 2 ' 

differentiating implicitly, we find, 4x 3 + 8y -i = 

dy _ -4x 3 -4x 3 _ ±x 3 

=? * dx ~~ 8y 



8(±i/r^ f ) (l-x^) 1 / 2 



y' 


4 
2 

V 


I 


) 




K 


/ 


( 


-l 


f- 

-2 

-4 


^ 


y 



x 4 + 4y*- - 1 



76. (x - 2) 2 + y 2 = 4: 

(a) y = ± v/4 - (x - 2) 2 

=> g = ± I (4 - (x - 2) 2 r 1/2 (-2(x - 2)) 
= [4 _^ x _~ 2 ^i/ 2 ; differentiating implicitly, 

-2(x-2) 



2(x-2) + 2y|=0^ g 



-(x-2) 



-(x-2) 



2y 
±(x-2) 



± [4 - (x - 2) 2 ] 1/2 [4 - (x - 2) 2 ] 1/2 



(b) 




(x-2) 2 +y 2 = 4 



77-84. Example CAS commands: 
Maple : 

ql := x A 3-x*y+y A 3 = 7; 

pt:=[x=2,y=l]; 

pi := implicitplot( ql, x=-3..3, y=-3..3 ): 

pi; 



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168 Chapter 3 Differentiation 

eval( ql, pt ); 

q2 := implicitdiff( ql, y, x ); 

m := eval( q2, pt ); 

tan_line := y = 1 + m*(x-2); 

p2 := implicitplot( tan_line, x=-5..5, y=-5..5, color=green ): 

p3 := pointplot( eval([x,y],pt), color=blue ): 

display( [pl,p2,p3], ="Section 3.6 #77(c)" ); 
Mathematica : (functions and xO may vary): 
Note use of double equal sign (logic statement) in definition of eqn and tanline. 

«Graphics" ImplicifPlof 

Clear[x, y] 

{xO, y0}={ l,7i74}; 

eqn=x + Tan[y/x]==2; 

ImplicitPlot[eqn,{ x, xO - 3, xO + 3),{y, yO - 3, y0+ 3j] 

eqn/.{x — » xO, y — » yO} 

eqn/.{ y -> y[x]} 

D[%, x] 

Solve[%, y'[x]] 

slope=y'[x]/.First[%] 

m=slope/.{x — * xO, y[x] — > yO) 

tanline=y==yO + m (x — xO) 

ImplicitPlot[{ eqn, tanline}, {x, xO - 3, xO + 3},{y, yO - 3, yO + 3}] 

3.7 RELATED RATES 



1. A = 7rr 2 =>• ^ = 27rr| 



2. S = 47rr 2 



dS q„~ dr 



3. (a) V = 7rr 2 h 
(c) V = 7rr 2 h 



dy _ _ r 2 dh 

dt ~~ ;11 dt 

dV „J1 dh i o^^U dr 

w = tit s + 27rrh a 



(b) V = 7rr 2 h 



f = 27rrh£ 



4. (a) V = ±7rr 2 h 



(c) 



_ = 1—2 dh 
dt 3 dt 



dY _ l_ r 2 dh 
dt ~~ t" 1 * 



dV 



3' 

vrrhl 



dt 



(b) V = i7rr 2 h 



dV 2_.v, dr 
^F = S 7 ™ dt 



5. (a) ^ = 1 volt/sec 
(0 f =R(f)+l(f 

. .. jn l r . ii / i 



dR _ 1 /dV 



(b) gj = — 5 amp/sec 



v ' dt 
1 /dV _ tj dl\ . dR _ 1 /dV _ V dl\ 

- u, - ui u, I I dt R dJ =? dt I V dt I dJ 

( d ) W = 2 t 1 _ f (~ I)] = (I) ( 3 ) = I ° hms / sec > R is increasing 



6. (a) P = RI 2 => f =I 2 f +2RI^ 



(b) P = RI 2 =* = f = I 2 § -r 

7. (a) s = ^?T7 = (x 2 + y 2 ) V2 
(b) s = ^Tf = (x 2 + y 2 ) 1/2 



9 RT dI . dR _ _ 2RI dl _ _ 2jf 
dt "T" ZK ' i dt ^ dt — I 2 dt ~ 



(c) s = a/x 2 + y 2 => s 2 = x 2 + y 2 
8. (a) s = yjx 2 + y 2 + z 2 =4> s 2 = x 2 + y 2 



2M dl _ _ 2(f) dl _ 2P dl 

I 2 dt I 2 dt I 3 dt 

ds x dx 

dt ~~ ,/x 2 + y 2 dt 

ds x dx i y dy 

h ~ V x2 + y 2 dt \/x 2 + y 2 dt 

2s | =2 x| + 2y| =► 2s-0 = 2x| 



dt 



2y| 



+ z 2 =* 2s|=2x£ + 2yf - 



-2z — 

zz dt 



dx 
dt 



y dy 
x dt 



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Section 3.7 Related Rates 169 



ds 
dt 



x dx 

\/x 2 + y 2 + z 2 dt 



(b) From part (a) with 2* = 

(c) From part (a) with % = 



y dy , z dz 

7x2+^2+12 dt "•" ^/ x 2 + y2 + z 2 dt 

_s. ds _ _y dy i 



dt 



= 2xf 



y X 2 + y2 + z 2 dt T v / x 2 + y2 + z 2 dt 

2y 



dy 
dt 



2z — 

""■ dt 



dx , y dy , z dz 
dt ' x dt ' x dt 



9. (a) A = i ab sin 9 
(c) A = \ ab sin 9 



^ = 1 ab cos 9 f 
^ = Iabcos0f 



(b) A = i ab sin 9 



| b sin i 



da 
dt 



db 



^ = 1 ab cos 9 § + 1 b sin (9 f 

dt 2 dt 2 dt 



10. Given A = tit 2 , g 



0.01 cm/sec, and r = 50 cm. Since ^ = 27rr || , then 



dAI 
dt I 





= 


7r cm 


7min. 










1. 


Given % 

dt 


= — 2 cm/sec 


dw 
' dt 


= 2 cm/sec, 


1 = 




(a) 


A = 


Av => ^ = 


/? dw 
* dt 


+ w£ 


=^ 


dA 
dt 




(b) 
(c) 


P = 
D = 


2£ + 2w => 


dP 

dt 

(w 2 


2^ + 2 


dw 

dt 






Vw 2 + P- = 


dU 

dt 



2*(50)Ug) 



12(2) + 5(— 2) = 14 cm 2 /sec, increasing 
2(— 2) + 2(2) = cm/sec, constant 

i(w 2 + f 2 r 1/2 (2w^ + 2/f) 



dD 
dt 



. 5X21+02X^2) _ _ M cm/seC; decreasing 



v / 25+~T44 



Vw 2 + < 



12. (a) V = xyz => f = yz | + xz | 



*y| 



dS 
dt 



dt I (4,3,2) 

(2y + 2z) | + (2x + 2z) % + (2x + 2y) d| 



dt _ J" dt 

(b) S = 2xy + 2xz + 2yz 

=* f 1 ,4,3,2, = (^X 1 ) + ( 12 )(- 2 ) + (14)(1) = m 2 /sec 

(c) I = v/x 2 + y 2 + z 2 = (x 2 + y 2 + z 2 ) 1/2 - «" - 



(3)(2)(1) + (4)(2)(-2) + (4)(3)(1) = 2 m 3 /sec 

dy i /"1„ i 0,A dz 



dx 



dy 



dj| 

dt I (4,3,2) 



'29 



dt y/x 2 + y 2 + z 2 dt ^/x 2 + y 2 + z 2 dt 

W+(^)(- 2 ) + fe)( 1 ) = 0m/SeC 



dz 



v / x 2 + y 2 + z 2 dt 



13. Given: $ = 5 ft/sec, the ladder is 13 ft long, and x = 12, y = 5 at the instant of time 



(a) Since x 2 + y 2 



169 



it 

dt 



x dx 
y dt 



(5) 



12 ft/sec, the ladder is sliding down the wall 



|xy 



dA 
dt 



(b) The area of the triangle formed by the ladder and walls is A 
is changing at \ [12( 12) + 5(5)] = - x -f = -59.5 ft 2 /sec. 

(c) cos0=^ => -sinflf = i-| =* f = _ n ^.^ = _(i) ( 5) = -irad/ S ec 



D(*f +yf)-Thear ea 



14. s 2 



= y 2 + X 2 = 

-614 knots 



2s 



2x 



dx 



2y£ 



d§ = I ( x dx , dy\ 
dt s \^ A dt x J dt ) 



ds 
dt 



7^ [5(-442) + 12(-481)] 



15. Let s represent the distance between the girl and the kite and x represents the horizontal distance between the 

,2 _ n.r\(\\2 i „2 _^ ds _ x dx _ 400(25) 

500 



girl and kite 



(300) 2 



ds 
dt 



x dx 
s dt 



20 ft/sec. 



16. When the diameter is 3.8 in., the radius is 1.9 in. and 



3000 



in/min. Also V = 67rr 



dV 
dt 



12tt(1.9) i 



3000 1 



0.00767T. The volume is changing at about 0.0239 in 3 /min. 



dV 
dt 



127T1- 



17. V = \ ?rr 2 h, h 



(a) 

(b) r 



dh I 

dt I h=4 

4h 
3 



= I (2r) = 

w) do) 

dr _ 4 dh 
dt — 3 dt 



,2, 



h(frh 



167rh 3 

27 



90 

256ir 



4 / 90 

3 V 2567T 



r= f => V 

0.1119 m/sec= 11.19 cm/sec 

0.1492 m/sec = 14.92 cm/sec 



dV _ 167rh 2 dh 
dt 9 dt 



15 

32tt 



Copyright (c) 2006 Pearson Education 




170 Chapter 3 Differentiation 



18. (a) V= |?rr 2 handr 



15h 



V 



1 _ M5hW 
3^1 — ) h 



75;rh 3 
4 



(b) r 



-0.0113 m/min = —1.13 cm/min 

15h . dr _ 15 dh . dr I _ / 15 \ / -8 \ _ -4 



2 dt 



dt I h=5 



2 ) \215t\) 



15-tt 



dV _ 2257rh 2 dh 
dt 4 dt 



0.0849 m/sec 



dh I 4(-50) -8 

dt I h=5 225tt(5) 2 225tt 

-8.49 cm/sec 



, -l 



19. (a) V=fy 2 (3R-y) => f = f [2y(3R - y) + y 2 (-l)] f =► f = [f (6Ry - 3y 2 )p f => at R = 13 and 
v " w "■" l " ,v " d ' " -jjj(-6) = £■= m/min 

(b) The hemisphere is on the circle r 2 + (13 — y) 2 = 169 

(c) r = (26y - y 2 ) 1/2 => $ = § (26y - y 2 r 1/2 (26 - 2y) £ = f 



_ -l 

144jt v "'' ~~ 24?r 

.2_,_na_,A2 _ iAQ -v r= v /26y - y 2 m 



13-8 



v/26y - y 2 dt 



dt I y=s ^26-8 - 64 V 24tt / 



gi m/min 



20. If V = | tit 3 , S = 4?rr 2 , and <^ = kS = 4k?rr 2 , then ^ = 47rr 2 | => 4k7rr 2 = 4m 2 | => | = k, a constant. 



Therefore, the radius is increasing at a constant rate. 



21. IfV= | Trr 3 , r = 5, and 



100tt ft 3 /min, then ^ = 47rr 2 | 



dr 
dt 



1 ft/min. Then S = 47rr 2 



dS 



87rr $ = 87t(5)(1) = 407r ft 2 /min, the rate at which the surface area is increasing. 



22. Let s represent the length of the rope and x the horizontal distance of the boat from the dock. 



(a) We have s 2 = x 2 + 36 = 
— I = IP (_2) 

dt I s=io Vl0 2 -36 V ' 

(b) cos 8 = f =4> - sin 8 % 

(-2) 



dt io 2 (^) 



dx s ds 


s 




ds 


dt x dt 


\/s 2 - 


36 


dt ' 


-2.5 ft/sec. 








6 di , 
r 2 dt ^ 


dfl 
dt — 




6 


r 2 


sin 8 


- J; rad/sec 









■i . Therefore, the boat is approaching the dock at 



. Thus, r = 10, x = 8, and sin 8 = -^ 



23. Let s represent the distance between the bicycle and balloon, h the height of the balloon and x the horizontal 
distance between the balloon and the bicycle. The relationship between the variables is s 2 = h 2 + x 2 



ds 
dt 



1 A, dh 



dx'i 



h f t + x dt 



ds 

dt 85 



,\ [68(1) + 51(17)] = lift/sec. 



24. (a) Let h be the height of the coffee in the pot. Since the radius of the pot is 3, the volume of the coffee is 



V = 97rh => 2V = 9n dh ^ the rate th coffee is risin ■ dh = l dv = io in/min 

dt dt ° dt 9tt dt 97T 



(b) Let h be the height of the coffee in the pot. From the figure, the radius of the filter r : 



jrh 3 



the volume of the filter. The rate the coffee is falling is 



dh _ 4 dV 
dt 7rh 2 dt 



4 

25:, 



=> V = 1 7rr 2 h 
(-10) = -± in/min. 



25. y = QD 



dy 



D 



-1 dQ 

dt 



q D -2 dD = i_ (Q) _ M3_ ( _ 2) = 4^_ L/min ^ increasing about 0.2772 L/min 



26. (a) 



dc 
dt 



(3x 2 - 12x + 15) § = (3(2) 2 - 12(2) + 15) (0.1) = 0.3. 



dr 



dx 



9(0.1) = 0.9, ^ 



0.9 - 0.3 = 0.6 



(b) | = (3x 2 - 12x - 45x- 2 ) | = (3(1.5) 2 - 12(1.5) - 45(1.5r 2 ) (0.05) = -1.5625, | = 70 f = 70(0.05) = 3.5, 
3.5 -(-1.5625) = 5.0625 



dp 
dt 



27. Let P(x, y) represent a point on the curve y = x 2 and 8 the angle of inclination of a line containing P and the 



origin. Consequently, tan 8 



tan i 



— = x 



sec 



2 a dfl _ dx 
dt ~~ dt 



cos 2 8 f . Since f = 10 m/sec 



and cos 8\ 



.=3 y2+x 2 



3 2 
9 2 +3 2 



■k , we have 



1 rad/sec. 



28. y = (-x) 1 / 2 and tan 8 



tan 6 



(-x) 1 - 



sec 2 f? 



x 2 dt 



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Section 3 .7 Related Rates 1 7 1 



dfl 
dt 



EB_ 



m 



(-8) 



(cos 2 (9) (f ) .Now, tan i 



rad/sec. 



cos I 



2 

v/5 



COS 2 fl = f 



Then 



29. The distance from the origin is s = y/x 



ds I 

dt I (5,12) 



(x 2 + y 2 )- 1/2 (2xf +2y£) 



(5.12) 



y 2 and we wish to find 

= < 5 X-l) + (12)(-5) = _ 5 ^ 
V25 + 144 



30. When s represents the length of the shadow and x the distance of the man from the streetlight, then s 



dl 
dt 



ds I dx 
dt ' dt 



(a) If I represents the distance of the tip of the shadow from the streetlight, then I = s + x 

(which is velocity not speed) => |atl = ll^t+^tl = llll^l = 5l _5 l = 8 ft/ sec > tne speed the tip of the 
shadow is moving along the ground. 

(b) 37 = I ^ = | ( — 5) = —3 ft/sec, so the length of the shadow is decreasing at a rate of 3 ft/sec. 



31. 



Let s = 16t 2 represent the distance the ball has fallen, 
h the distance between the ball and the ground, and I 
the distance between the shadow and the point directly 
beneath the ball. Accordingly, s + h = 50 and since 
the triangle LOQ and triangle PRQ are similar we have 



I 



30h . 

" 50 -h =? * 

1500 _ on 
16t 2 JU 



50 - 16t 2 and I 



30(50- 16t 2 ) 
50-(50-16t 2 ) 



dl 

dt 



1500 
8t? 



dl I 
dt I 



1500 ft/sec. 



Ball at time t = 



sec later 




SP J a A0 _ 1 ds 
sec o d( — 132 d( 



32. Let s = distance of car from foot of perpendicular in the textbook diagram =4> tan 9 — jf* 

-264 and 9 = =>• % = —2 rad/sec. A half second later the car has traveled 132 ft 

£ .2 



cos 2 8 ds . ds 
132 dt ' dt 



right of the perpendicular =>■ \8\ = |, cos 9 



i and f 

2 dt 



264 (since s increases) 



1(264) 



1 rad/sec. 



33. The volume of the ice is V 



7T4 d 



dV 
dt 



471T 



.2 dr 



— I 
dt I r=6 



thickness of the ice is decreasing at =f- in/min. The surface area is S = 47rr 2 



wJ- in./min when ^ 

12tt dt 

dS o^-„ dr * dS I 



-10in 3 /min, the 



10 



— y in/min, the outer surface area of the ice is decreasing at y in J /min 



34. Let s represent the horizontal distance between the car and plane while r is the line-of-sight distance between 
the car and plane => 9 + s 2 = r 2 => % = -^= % =^ %\ = -2- (-160) = -200 mph 

r dt ,y r 2_9 dt dt I r=5 ,^16 v ' r 

=> speed of plane + speed of car = 200 mph =>■ the speed of the car is 80 mph. 



35. When x represents the length of the shadow, then tan 9 = — =>• sec 2 9 ^ 



We are given that 



0.27° 



I dx I 

I dt I 



-x 2 sec 2 dfl 
dt 



80 



2^g rad/min. At x 
= 3| ft/min » 



80 . 

x 

= 60, cos 9 = 
0.589 ft/min 



80 dx 

x 2 dt 



7.1 in./min. 



dx 
dt 



at A 



sec 2 9 



M 



36. Let A represent the side opposite 9 and B represent the side adjacent 9. tan 9 = g —r ow w d ,, 
• - - * -10mandB = 20mwehavecosc9 =I ^ = -2=andf =[(i)(-2)-(i|(l))](| 

-^/sec w -67sec 



l dA 
B dt 



A_ dB 

B 2 dt 



'^1 1_\ /4^ 

v 10 40 J V5/ 



— -Tfj rad/sec 



37. Let x represent distance of the player from second base and s the distance to third base. Then 



dx 
dt 



16 ft/sec 



(a) s 2 



8100 



2s - 

-" dt 



2x ^ 
zx dt 



ds 
dt 



x dx 
s dt 



When the player is 30 ft from first base, x = 60 



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172 Chapter 3 Differentiation 



13 and % = -%- (-16) = ^2= « -8.875 ft/sec 

dt 30a/13 V13 

= &■%. Therefore, x = 60 and s = 30x/l3 

sx dt ' v 



(b) cos 0i = ? => -sin^f = -f -| => f 

-^ (=£-\ = =£ rad/sec; sin 2 



ds . dfli _ 90 ds _ _ 

" " ' dt — sx 



s 2 sin #i 



cWi _ 



2 = 2p => cos0 2 f 



dt (30/b) (60) 

Therefore, x = 60 and s = 30v/l3 => ^ = ^ rad/sec. 



d«2 _ _ 90 . ds . dfl2_ -90 
dt s 2 ' dt dt s 2 cos 9 2 



-90 ds 
dt 



= ^0 Got) (-15) 

= (x^floo)f =► x l^ f = I rad/sec 



= ou ana s = juy u =?• "dt = 65 raa/s 

\sJ ' \dt) Vs 2 MdJ Vx 2 + 8100-' dt 



lim 4L 

x -> dt 



Cc) dJl _ 90 . ds _ 90 . (x^ 

v ' dt s 2 sinfli dt ( s2 ';) ^ s ' vat/ \ s* / \ at / \x^ + siuu/ at x — > at 

„. d02 _ -90 ds _ /"^9o\ /x\ fdx\ _ ( = W\ /dx\ 

" ' isfl 2 ' dt ~~ \^--J VsJ VdJ ~~ V s 2 ) VdJ 



-15) = -±rad/sec;f = ? 



COS #2 



38. Let a represent the distance between point O and ship A, b the distance between point O and ship B, and 
D the distance between the ships. By the Law of Cosines, D 2 = a 2 -f-b 2 2ab cos 120° 



J(> " 2D [2a|+2b^ + a db 



dt 2D 



dt ' " dt 



b | ] . When a = 5, | = 14, b = 3, and f = 21, then ^ - 4 " 



2D 



where D = 7. The ships are moving ^ = 29.5 knots apart. 



3.8 LINEARIZATION AND DIFFERENTIALS 

1. f(x) = x 3 - 2x + 3 =>■ f'(x) = 3x 2 - 2 =4> L(x) = f'(2)(x - 2) + f(2) = 10(x - 2) + 7 =>• L(x) = lOx - 13 at x = 2 



2. f(x) = Vx 2 + 9 = (x 2 + 9) 1/2 =>• f'(x) = (I) (x 2 + 9) 1/2 (2x) 
= - | (x + 4) + 5 => L(x) = -|x+|atx=-4 



V^ 2 + 9 



L(x) = f '(-4)(x + 4) + f(-4) 



3. f(x) = x + i => f'(x) = 1 - x- 2 => L(x) = f(l) + f'(l)(x - 1) = 2 + 0(x - 1) = 2 

4. f(x) = xV3 ^ f ( X ) = ^L_ => L (x) = f'(-8)(x - (-8)) + f(-8) = i (x + 8) - 2 => L(x) = i x - f 

5. f(x) = x 2 + 2x =>■ f'(x) = 2x + 2 =>■ L(x) = f '(0)(x - 0) + f(0) = 2(x - 0) + => L(x) = 2x at x = 

6. f(x) = x" 1 =>■ f'(x) = -x" 2 => L(x) = f'(l)(x - 1) + f(l) = (-l)(x - 1) + 1 =4> L(x) = -x + 2 atx = 1 

7. f(x) = 2x 2 + 4x - 3 =>• f'(x) = 4x + 4 => L(x) = f'(-l)(x+ 1) + f(-l) = 0(x + 1) + (-5) => L(x) = -5atx= -1 

8. f(x) = 1 + x => f (x) = 1 => L(x) = f'(8)(x - 8) + f(8) = l(x - 8) + 9 => L(x) = x + 1 at x = 8 

9. f(x) = \/x = x 1 / 3 => f'(x) = (i) x- 2 / 3 => L(x) = f'(8)(x - 8) + f(8) = i (x - 8) + 2 => L(x) =^x+|atx = 8 



10. f(x) 



x+l 



f'(x) 



(l)(x+l)-(l)(x) 



(x+l) 2 



(x+l) 2 



L(X) = f'(l)(x - 1) + f(l) = 1 (X - 1) + 1 



=> L(x) =i x +iatx=l 



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Section 3.8 Linearization and Differentials 173 



11. f(x) = sin x =>■ f (x) = cos x 

(a) L(x) = f (0)(x - 0) + f(0) = l(x - 0) + 

=> L(x) = x at x = 

(b) L(X) = f (7T)(X - 7r) + f(7T) = (- 1)(X - 7T) + 

=> L(x) = 7T — xatx = 7r 



L(x) = TT - X 




L(x)=x/ " 2 f(x)=sinx 



12. f(x) = cos x =^ f'(x) = —sin x 

(a) L(x) = f (0)(x - 0) + f(0) = 0(x - 0) + 1 

=> L(x) = 1 at x = 

(b) L(x) = f'(-f)(x+|)+f(-f) 

= -(-1) (x + f ) + => L(x) = x + f 
at x = - 5 



2 

L(x)=l / 


^J-(x)=x+f 


s 


^\ 


-T, y -\ 
' -2 


1 s ^2 3 
f(x) =C0S X 



13. f(x) = sec x =>■ f (x) = sec x tan x 

(a) L(x) = f (0)(x - 0) + f(0) = 0(x - 0) + 1 
=> L(x) = 1 at x = 



(b) L(x) = f (- 
= -2>/3(x 

at x = - 5 



fM x +i 



•fi 



2 => L(x) = 2 - 2V3 (x + f ) 




L(x) =2-2/3"(x+i) 0.5 



■*~f (x) = sec x 



14. f(x) = tan x => f (x) = sec 2 x 

(a) L(x) = f (0)(x - 0) + f(0) = l(x - 0) + = x 

=> L(x) = x at x = 

(b) L(x) = f (|) (x - |) + f (|) = 2 (x - |) + 1 



=4> L(x) = 1 + 2 (x 



^jatx= % 



f(x) = tan x 




X L(x)=l *2(x-J) 



\k-l 



15. f (x) = k(l + x) K \ We have f(0) = 1 and f (0) = k. L(x) = f(0) + f (0)(x - 0) = 1 +k(x - 0) = 1 + kx 



16. (a) f(x) = (1 - x) G = [1 + (-x)]° w 1 + 6(-x) = 1 - 6x 

(b) f(x) = r ^=2[l + (-x)]" 1 «2[l + (-l)(-x)] =2 + 2x 

(c) f(x) = (l + x)- 1/2 «l+(-i)x=l-| 

1 fi 

(d) f(x) = y/T+** = y/2[ 1+ f ) wy/2(l + If) = 72(1+1 



(e) f(x) = (4 + 3x) 1/3 = 4^3(1 + I) 1 / 3 « 4 i/3(i + §f ) = 4^(1 + |) 



1X2 

2 2 

1 

:s 



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174 Chapter 3 Differentiation 



(f) f(x) = (1 - 



1 \2/3 
2 + x) 



' 2 + x, 



2/3 



3 V 2 + x ) 



6+3x 



17. (a) (1.0002) 5U = (1 + 0.0002) 5U « 1 + 50(0.0002) = 1 + .01 = 1.01 
(b) V 1.009 = (1 + 0.009) 1/3 « 1 + (i) (0.009) = 1 + 0.003 = 1.003 



18. f(x) = y/\ + 1 + sin x = (x + l) 1 / 2 + sin x => f (x) = (i) (x + l)" 1 / 2 + cos x => L,(x) = f (0)(x - 0) + f(0) 
= § (x - 0) + 1 => L, (x) = \ x + 1, the linearization of f(x); g(x) = y/x + 1 = (x + l) 1 / 2 => g'(x) 
= (i) (x + 1) _1/2 => L g (x) = g'(0)(x - 0) + g(0) = i (x - 0) + 1 =$► L g (x) = i x + 1, the linearization of g(x); 
h(x) = sin x =^ h'(x) = cos x =^ L h (x) = h'(0)(x - 0) + h(0) = (l)(x - 0) + => L h (x) = x, the linearization of 
h(x). L f (x) = L g (x) + L h (x) implies that the linearization of a sum is equal to the sum of the linearizations. 



19. 



y = x 3 - 3v/x = x 3 - 3X 1 / 2 => dy = (3x 2 - § x" 1 / 2 ) dx => dy = ^3x 2 - ^A 



dx 



20. y = x V / l-x 2 = x(l-x 2 ) 1/2 => dy= [(l)(l-x 2 ) 1/2 + (x)(±)(l-x 2 ) 1/2 (-2x)j dx 



1-x 2 ) 1/2 [(l-x 2 )-x 2 ]dx 



(l-2x 2 ) 
Vl-x 2 



dx 



2x 



21. y- — 

22. y = 



H„ _ f (2)(l+x 2 )-(2x)(2x) \ 
ay ~\ U+x 2 ) 2 J 



dX =^ dX 



2v^ 2xV2 , /x- 1 /2(3(l+x 1 /2))_2x 1 /2(|x- 1 /2)\ ix , _ , , , 

3(iT7^ = WTxVT) => dy = ( 9(1 ^ xl/2f J j dx = > , , ox 



3X- 1 / 2 + 3-3 
9(l+xV2) 2 



dy = — ^v^ — f^ dx 

3 3^(1 + ^) 2 



23. 2y 3 / 2 + xy - x = => 3y 1/2 dy + y dx + x dy - dx = =>■ (3y 1/2 + x) dy = (1 - y) dx => dy ■- 

24. xy 2 - 4x 3 / 2 - y = => y 2 dx + 2xy dy - 6X 1 / 2 dx - dy = =4> (2xy - 1) dy = (6X 1 / 2 - y 2 ) dx 

^dy=^dx 



i-y 

3^+x 



dx 



25. y = sin (5,/x) = sin (5X 1 / 2 ) => dy = (cos (5X 1 / 2 )) (| x" 1 / 2 ) dx => dy 



5 cos (5^/xj 



dx 



26. y = cos (x 2 ) =>■ dy = [—sin (x 2 )] (2x) dx = — 2x sin (x 2 ) dx 

27. y = 4 tan Nf\ => dy = 4 (sec 2 (f )) (x 2 ) dx ^ dy = 4x 2 sec 2 (jf) dx 

28. y = sec (x 2 - 1) =4> dy = [sec (x 2 - 1) tan (x 2 - 1)] (2x) dx = 2x [sec (x 2 - 1) tan (x 2 - 1)] dx 

29. y = 3 esc (l - 2^/x) = 3 esc (l - 2x J / 2 ) => dy = 3 (-esc (l - 2X 1 / 2 )) cot (l - 2X 1 / 2 ) (-x^ 1 / 2 ) dx 

=> dy = 4^ esc (l - 2yfx) cot (l - 2-^x) dx 



30. y = 2 cot (-±) = 2 cot (x^ 1 / 2 ) => dy = -2 esc 2 (x^ 1 / 2 ) (- i) (x~ 3 / 2 ) dx =>- dy = -^ esc 2 (-^) 

31. f(x) = x 2 + 2x, x = 1, dx = 0.1 => f'(x) = 2x + 2 

(a) Af = f(x + dx) - f(x ) = f(l.l) - f(l) = 3.41 - 3 = 0.41 

(b) df = f'(x ) dx = [2(1) + 2](0.1) = 0.4 

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dx 



Section 3.8 Linearization and Differentials 175 



(c) |Af -df I = |0.41 -0.4| =0.01 

32. f(x) = 2x 2 + 4x - 3, x = -1, dx = 0.1 => f'(x) = 4x + 4 

(a) Af = f(x + dx) - f(x ) = f(-.9) - f(-l) = .02 

(b) df=f'(x )dx=[4(-l) + 4](.l) = 

(c) |Af-df| = |.02-0| = .02 

33. f(x) = x 3 - x, x = 1, dx = 0.1 =>■ f'(x) = 3x 2 - 1 

(a) Af = f(x + dx) - f(x ) = f(l.l) - f(l) = .231 

(b) df=f'(x )dx=[3(l) 2 -l](.l) = .2 

(c) |Af-df| = | .231 -.2| = .031 

34. f(x) = x 4 , x = 1, dx = 0.1 => f (x) = 4x 3 

(a) Af = f(x + dx) - f(x ) = f(l.l) - f(l) = .4641 

(b) df = f'fo) dx = 4(l) 3 (.l) = .4 

(c) |Af - df | = |.4641 - .4 1 = .0641 

35. f(x) = x -1 , x = 0.5, dx = 0.1 =>■ f'(x) = -x~ 2 

(a) Af = f(x + dx) - f(x ) = f(.6) - f(.5) = - | 

(b) df=f'(x )dx = (-4)(i) = -§ 

(c) |Af-df| = |-| + || = i 

36. f(x) = x 3 - 2x + 3, x = 2, dx = 0.1 => f'(x) = 3x 2 - 2 

(a) Af = f(x + dx) - f(x ) = f(2.1) - f(2) =1.061 

(b) df = f'(x ) dx = (10X0.10) = 1 

(c) |Af-df| = 1 1.061 - II = .061 



37. V = | ?rr 3 => dV = 47rr 2 dr 



38. V = x 3 => dV = 3x 2 dx 



39. S = 6x 2 =>• dS = 12x dx 



40. S = m v / r y Th 2 = m(r 2 + h 2 ) 1/2 ,hr— - -■ ds - - '- 2 ' ^ 1/2 ^ — - .« j ^-i - 



dS _ 7T (r 2 + h 2 ) + m 2 
dr Vr 2 +h 2 



constant =4> ^ = 7r (r 2 + h 2 ) ' + 7rr • r (r 2 + h 2 )~ 

2. 

W+ b2 



dS = 7r( ^ + h2) dr, h constant 



41. V = 7rr 2 h, height constant =>• dV = 27rroh dr 



42. S = 2?rrh => dS = 2?rr dh 



43. Given r = 2 m, dr = .02 m 

(a) A = tit 2 =>■ dA = 2?rr dr = 2tt(2)(.02) = .08tt m 2 

(b) (ff)(100%) = 2% 



44. C = 27rr and dC = 2 in. 
= 2tt(5)(±) = 10 in. 2 



=^> dC = 2?r dr =>• dr 



the diameter grew about - in.; A = 7rr 2 =^> dA = 27rr dr 



45. The volume of a cylinder is V = 7rr 2 h. When h is held fixed, we have ^ = 27rrh, and so dV = 27rrh dr. For h = 30 in. 
r = 6 in., and dr = 0.5 in., the volume of the material in the shell is approximately dV = 27rrh dr = 27r(6)(30)(0.5) 
= 180tt« 565.5 in 3 . 



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176 Chapter 3 Differentiation 

46. Let 8 = angle of elevation and h = height of building. Then h = 30tan 6, so dh = 30sec 2 (9 A9. We want |dh| < 0.04h, 



which gives: |30sec 2 6» d0| < 0.04|30tan6»| 



,.fiu 



\m< 



0.04sin 



|d0| < 0.04sin 6 cos 9 =>• |d0| < 0.04sin ff cos 



5jt 
12 



= 0.01 radian. The angle should be measured with an error of less than 0.01 radian (or approximatley 0.57 degrees), 
which is a percentage error of approximately 0.76%. 



47. V = Trh 3 => dV = 3?rh 2 dh; recall that AV « dV. Then |AV| < (1%)(V) 



(l)(Th 3 ) 
100 



IdVl < 



(l)(vrh 3 ) 



|37rh 2 dhl < 



(l)(7rh 3 ) 
100 



|dh| < joy h = (5 %) h. Therefore the greatest tolerated error in the measurement 



of his i%. 



7T(|) 2 h 



7rD?h 



and h = 10 



48. (a) Let D ; represent the inside diameter. Then V = 7rr h — v ,, •■ 

dV = 57TD; dDi. Recall that AV « dV. We want | AV| < (1%)(V) => |dV| < 



. 100 J 



> V 

2 ' 



5jrD'f 
~ 2~ 

_ 7TD[ 

~ 40 



5ttD, dD ; < 



;vD- 



dDi 



< 200. The inside diameter must be measured to within 0.5%. 



40 "^ Di 

(b) Let D e represent the exterior diameter, h the height and S the area of the painted surface. S = 7rD e h 



dS = TrhdD, 



dS 
S 



^p . Thus for small changes in exterior diameter, the approximate percentage change in the exterior diameter 



is equal to the approximate percentage change in the area painted, and to estimate the amount of paint required to 
within 5%, the tanks's exterior diameter must be measured to within 5%. 



49. V = 7rr 2 h, h is constant => dV = 27rrh dr; recall that AV « dV. We want I AVI < 



I27rrh drl < — =^> Idrl < - L - 
|z,/uuui| \ 100() =? |ui| \ 2()00 



(.05%)r 



1000 



V 



IdVl < 



7rr 2 h 
1000 



a .05% variation in the radius can be tolerated. 



50. Volume = (x + Ax) 3 



3x 2 (Ax) + 3x(Ax) 2 + (Ax) 3 




51. W = a+!?=a + bg" 1 => dW = -bg~ 2 dg 



dW m „„ 
dW t „ h 



bdg 
"(5.2)2 



bdg ' 
"(32)2, 



'jk\ = 37.87, so a change of 



gravity on the moon has about 38 times the effect that a change of the same magnitude has on Earth. 

I try 

52. (a) T = 2tt (|) =» dT = 2tt\/l (- \ g" 3 / 2 ) dg = -n^g-V 2 dg 

(b) If g increases, then dg > =4> dT < 0. The period T decreases and the clock ticks more frequently. Both 
the pendulum speed and clock speed increase. 

(c) 0.001 = -TT^/lOO (980~ 3 / 2 ) dg =>• dg w -0.977 cm/sec 2 => the new g w 979 cm/sec 2 



53. The error in measurement dx = (1%)(10) = 0.1 cm; V = x 3 => dV = 3x 2 dx = 3(10) 2 (0.1) = 30 cm 3 => the 

' 30 \ 
,1000 7 



percentage error in the volume calculation is (y^) (100%) = 3% 



Copyright (c) 2006 Pearson Education 




Section 3.8 Linearization and Differentials 177 



54. A = s 2 => dA = 2s ds; recall that AA « dA. Then |AA| < (2%)A 



m = 51) ^ l dA l - 50 



|2sds|<fg 



Idsl < 



(2s)(50) 100 



(1%) s => the error must be no more than 1% of the true value. 



55. Given D = 100 cm, dD = 1 cm, V = \ tt ( D ^ 3 - nti> 



3 " V 2 . 



dV= ^D 2 dD = | (100) 2 (1) 



10 4 ji 



Then ^ (100%) 



10 6 ir 
6 



(10 2 %) 



infij 



% = 3% 



56. V=|7rr 3 = |7r(§] 



dV = 2f? dD; recall that AV w dV. Then | AV| < (3%)V 



3 ' 
100, 



(*) 



-D 3 

201) 



|HVl < — => — dD < ^ 

l uv ! — 200 ^ 2 "^ — 200 



dD|<y^ = (l%)D =>■ the allowable percentage error in 



measuring the diameter is 1%. 



57. A 5% error in measuring t => dt = (5%)t = ^ . Then s = 16t 2 =4> ds = 32t dt = 32t (4 



32t 2 _ 16t 2 
v20; — 20 ~~ 10 



.10) 



= (10%)s =>• a 10% error in the calculation of s. 
58. From Example 8 we have y=4y. An increase of 12.5% in r will give a 50% increase in V. 



59. lim 

x^0 



'1+x Vl+0 



1 + 1 



60. lim ^nx 

x^0 x 



lim (5&i) (-M 

„ y Q V X / V COS X / 



(D(l) = 1 



61. E(x) = f(x) - g(x) => E(x) = f(x) - m(x - a) - c. Then E(a) = =>■ f (a) - m(a - a) - c = => c = f(a). Next 
we calculate m: lim ^ = => lim f(x) ~ m(x ~ a) ~ c = => lim [ f(x) ~ f(a) - ml = (since c = f(a)) 

x^a x-a x^a x-a x ^ a ^ x -a J v ^ " 

=> f'(a) — m = => m = f '(a). Therefore, g(x) = m(x — a) + c = f '(a)(x — a) + f(a) is the linear approximation, 
as claimed. 



62. (a) i. Q(a) = f(a) implies thatb = f(a). 

ii. Since Q'(x) = bj + 2b 2 (x - a), Q'(a) = f'(a) implies that bj = f'(a). 

iii. Since Q"(x) = 2b 2 , Q"(a) = f"(a) implies that b 2 = ^. 
In summary, b = f(a), bj = f'(a), and b 2 
(b) f(x) = (1 - x)- 1 

f'(x) = -l(l-x)- 2 (-l) = (l-x)- 2 
f"(x) = -2(1 - x)" 3 (-l) = 2(1 - x)~ 3 



fja) 
2 ' 



Since f(0) = 1, f'(0) = 1, and f"(0) = 2, the coefficients are b = 1, bj = 1, b 2 
approximation is Q(x) = 1 + x + x 2 . 



1 . The quadratic 



(c) 




As one zooms in, the two graphs quickly become 
indistinguishable. They appear to be identical. 



[-2.35, 2.35]by [-1.25, 3.25] 
(d) g(x) = x- 1 
g'(x) = -lx- 2 
g"(x) = 2x~ 3 
Since g(l) = 1, g'(l) = — 1, and g"(l) = 2 , the coefficients are bo = 1, bi 



-l,b 2 



1 . The quadratic 



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178 Chapter 3 Differentiation 



approximation is Q(x) = 1 — (x — 1) + (x — 1) 




As one zooms in, the two graphs quickly become 
indistinguishable. They appear to be identical. 



[-1.35,3.35]by[-1.25,3.25] 

(e) h(x) = (l+x) 1/2 
h'(x) = i(l + x)- 1/2 
h"(x) = -i(l+x)- 3 / 2 

Since h(0) = 1, h'(0) = \, and h"(0) 
approximation is Q(x) = 1 + | — y. 




-\ , the coefficients are bo = 1, bi = \,^i = -f = — |. The quadratic 



As one zooms in, the two graphs quickly become 
indistinguishable. They appear to be identical. 



[-1.35, 3.35]by [-1.25, 3.25] 
(f) The linearization of any differentiable function u(x) at x = a is L(x) = u(a) + u'(a)(x — a) = bo + bi(x — a), where 
bo and bi are the coefficients of the constant and linear terms of the quadratic approximation. Thus, the linearization 
for f(x) at x = is 1 + x; the linearization for g(x) at x = 1 is 1 — (x — 1) or 2 — x; and the linearization for h(x) at 
x = 0isl + |. 

63. (a) x = 1 







5rr2 














Y-*y\ 







9 0. 


95 j* 


l.jOS 1 


1 








i 


(b) x= l;m = 2.5, e 1 « 2.7 









9 0. 


95 1. 


*S 


1 








y.S/ 


































frrS 









x = 0;m= l,e°= 1 



-*T«i 



«.l -0-05 



-«;-9 



05 



l 
i 

Oil 



l;m = 0.3, e- 1 w 0.4 




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Section 3.8 Linearization and Differentials 179 



64. If f has a horizontal tangent at x = a, then f ' (a) = and the linearization of f at x = a is 
L(x) = f(a) + f'(a)(x — a) = f(a) + - (x — a) = f(a). The linearization is a constant. 



65. Find |v| when m = l.Olnio. m = 

=> |v| = c\/l - 55? => dv = c • 
dv= - 



t^T^ m A /l 



2\ x n?) V m3 / 



m = ^/I - K = ^ =■ 1 
dm, dm = O.Olmo =>• dv = 



v 2 _ m 2 
c 2 m 2 



mp 



m 2 VlOO^ 
1-1 



c-m* 



101 3 m 3 



1— #- 

\j 100 2 o 



'mo.) 
-100) 



1000 



101 3 i/l- 



100 2 

10 1 2 



0.69c. Body at rest => v = and v = v + dv 



0.69c. 



c* 1 



„, — 101 ^ 

m - M m , 



" 2 / 



66. (a) The successive square roots of 2 appear to converge to the number 1 . For tenth roots the convergence is more rapid, 
(b) Successive square roots of 0.5 also converge to 1. In fact, successive square roots of any positive number converge 
tol. 
A graph indicates what is going on: 




p- x 



0.5 1 1.5 2 

Starting on the line y = x, the successive square roots are found by moving to the graph of y 
the line y = x again. From any positive starting value x, the iterates converge to 1 . 



^/x and then across to 



67-70. Example CAS commands: 
Maple : 

with(plots): 

a: = l:f:=x->xA3+A- 2 2*x; 

plot(f(x), x=-1..2); 

diff(f(x),x); 

fp := unapply (",x); 

L:=x -> f(a) + fp(a)*(x - a); 

plot({f(x),L(x)),x=-1..2); 

err:=x -> abs(f(x) — L(x)); 

plot(err(x), x=— 1..2, title = #absolute error function*); 

err(— 1); 
Mathematica : (function, xl, x2, and a may vary): 

Clearff, x] 

{xl,x2} = {-l,2};a = 1; 



f[x_]:=x 



_ v 3 



2x 



Plot[f[x],{x,xl,x2}] 
lin[x_]=f[a] + f [a](x - a) 
Plot[{f[x],lin[x]},{x,xl,x2}] 
err[x_]=Abs[f[x] - lin[x]] 



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180 Chapter 3 Differentiation 

Plot[err[x],{x,xl,x2}] 
err//N 
After reviewing the error function, plot the error function and epsilon for differing values of epsilon (eps) and delta (del) 
eps = 0.5;del = 0.4 
Plot[{err[x], eps},{x, a — del, a + del}] 

CHAPTER 3 PRACTICE EXERCISES 



1. y = x 5 - 0.125x 2 + 0.25x =>• f = 5x 4 - 0.25x + 0.25 



2. y = 3 - 0.7x 3 + 0.3x 7 => 

3. y = X 3 - 3 (X 2 + 7T 2 ) => 

4. y = x 7 + Jlx - 



g = -2.1x 2 + 2.1x 6 



| = 3x 2 - 3(2x + 0) = 3x 2 - 6x = 3x(x - 2) 



^T =* %=T? + S> 



5. y = (x+ l) 2 (x 2 + 2x) 

= 2(x+ l)(2x 2 + 4x+ 1) 



dv 



(x + l) 2 (2x + 2) + (x 2 + 2x) (2(x + 1)) = 2(x + 1) [(x + l) 2 + x(x + 2)] 



6. y = (2x - 5)(4 - x)" 1 => g = (2x - 5)(-l)(4 - x)- 2 (-l) + (4 - x)- x (2) = (4 - x)- 2 [(2x - 5) + 2(4 - x) 
= 3(4 - x)-' 2 



7- y 



9 2 + sec 9 + l) 3 => t = 3 (6» 2 + sec 9 + \f{26 + sec 9 tan 0) 



8. y 



(_, _ cscj? _ fpY 

v 2 " v 



-■ ■ - 2 (- 



J> 



-)( 5 



CSC ( 

2 



| (esc 61 cot 9 - 6) 



9. s 



Vt ^ ds _ o+v^-^-^^) _ (i + v^)-yt 



i+Vt dt 



(i+^t) 



20(1 + 7^ 20(1 + ^ 



10. s 



i ^ ds= (>A-0(Q)-i(^) = _i 

v/t-l dt (v^-1) 2 2^(^-l) 2 



11. y = 2 tan 2 x — sec 2 x =>• gf = (4 tan x) (sec 2 x) — (2 sec x)(sec x tan x) = 2 sec 2 x tan x 



12. y = jjp-j — -tj- = esc 2 x — 2 esc x =>• ■£ = (2 esc x)(— esc x cot x) — 2(— esc x cot x) = (2 esc x cot x)(l — esc x) 



13. s = cos 4 (1 - 2t) => | = 4 cos 3 (1 - 2t)(-sin (1 - 2t))(-2) = 8 cos 3 (1 - 2t) sin (1 - 2t) 

14. s = cot 3 (|) * | = 3 cot 2 (|) (-esc 2 (f)) (f ) = | cot 2 (f) esc 2 (f) 

15. s = (sec t + tan t) 5 => ^ = 5(sec t + tan t) 4 (sec t tan t + sec 2 t) = 5(sec t)(sec t + tan t) 5 

16. s = csc 5 (l -t + 3t 2 ) =4> f t =5csc 4 (l -t+3t 2 )(-csc(l -t + 3t 2 )cot(l - t + 3t 2 )) (-1 + 6t) 

= -5(6t- l)csc 5 (l -t + 3t 2 )cot(l — t + 3t 2 ) 



17. r = JlB sin e = (29 sin (9) 1/2 => ^ = ± (26> sin 9)- l l 2 (29 cos 6> + 2 sin 0) = gco / se + sing 

ay l y 20 sin 

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Chapter 3 Practice Exercises 181 



18. r = 26> a/cos 9 = 29 (cos 9) 1 !' 2 

_ 2 cos 9-9 sin 9 



29 (1) (cos 6»)- 1 / 2 (-sin (9) + 2(cos 6>) 1/2 = =p^ + 2^fcos9 



/cos 9 



19. r = sin ^29 = sin(20) 1/2 => % = cos(29) 1 / 2 (1 (26T 1/2 (2)) = 52^ 



20. r = sin 



(0 + v^ 



(. + ^TT)(i + ^ TT ) = 2 ^cos(. + ^TT) 



i v2 re ? -^ dy - I v2/_ m , 2 „- t 2\ /=2 



21. y = 5X 2 esc = =>• 



x — esc £ cot 



2\ i\ 



X X / V x^ 



(csc -I I $ ■ 2xj = csc - cot - + x csc 



22. y = 2v/x" sin yfi => g = 2^ (cos ^ ( 0-) + (sin ^/x) (^) = cos y/i 



2\A 



2 \ / — , sin 

~ ! — COS \/X 



sA 



y^ 



23. y = x- 1 / 2 sec (2x) 2 



J) 



- 1 / 2 sec (2x) 2 tan(2x) 2 (2(2x) • 2) + sec (2x) 2 (- ± x" 3 / 2 ) 



= 8X 1 / 2 sec(2x) 2 tan(2x) 2 - \ x~ 3 / 2 sec(2x) 2 = \ x 1 / 2 sec(2x) 2 [16 tan(2x) 2 - x~ 2 ] or ^sec(2x) 2 [l6x 2 tan(2x) 2 - l] 
24. y = v/x csc (x + l) 3 = x 1 / 2 csc (x + l) 3 



(-csc(x+ l) 3 cot(x+ l) 3 )(3(x+ l) 2 )+csc(x+ l) 3 (ix^ 1 /' 2 ) 



dy _ Y l/2 
dx — A 

-3v/x(x+ l) 2 csc(x+ l) 3 cot(x+ l) 3 + csc ^ x + 1)3 = \ v /xcsc(x+ l) 3 [i - 6(x+ l) 2 cot(x+ l) 3 ] 



or A^csc(x + 1) 3 [1 - 6x(x + l) 2 cot(x + l) 3 ] 

25. y = 5cotx 2 => ^ = 5 (-csc 2 x 2 ) (2x) = -lOx csc 2 (x 2 ) 

26. y = x 2 cot 5x =$■ g| = x 2 (-csc 2 5x) (5) + (cot 5x)(2x) = -5x 2 csc 2 5x + 2x cot 5x 

27. y = x 2 sin 2 (2x 2 ) => g = x 2 (2 sin (2x 2 )) (cos (2x 2 )) (4x) + sin 2 (2x 2 ) (2x) = 8x 3 sin (2x 2 ) cos (2x 2 ) + 2x sin 2 (2x 2 ) 

28. y = x~ 2 sin 2 (x 3 ) =4> & = x~ 2 (2 sin (x 3 )) (cos (x 3 )) (3x 2 ) + sin 2 (x 3 ) (-2x~ 3 ) = 6 sin (x 3 ) cos (x 3 ) - 2x~ 3 sin 2 (x 3 ) 

on „ _ (JL_\- 2 -v ds _ o (^\- 3 f (t+l)(4)-(4t)(l) \ _ _ 9 /Jl\-3 4 _ _ (t+1) 

Zy - S— Vt+lJ ^ dt ~~ Z Vt+lJ ^ (t+1) 2 J — z Vt+lJ (t+1) 2 — 8t? 



30. s 



-l 

15(15t-l) 3 



i (15t - I)" 3 =► | = - i (-3)(15t - 1) 4 (15) = ^ 



I)" 1 



3i-y = (x4) =*! = 2(i&) 



vM , »+')(^)~(>A)(') _ (x+1)- 



2x 1 - x 



(x+1) 2 



(X+1)3 (X+1)3 



32 -y = {^) : 



Wi \ / fiyg+i^-fty^) ^ _ Vx(4 E ) 



dy _ 2 f 2 \A A 

dx V2V^+l/ I (2,/x+l) 2 



(2^+l) J (2^+l) J 



33. y 



x 2 + x 

X 2 



IN 1/2 dy 



-1/2 



2x 2 \\ + I 



34. y = 4x^x + ^ = 4x (x + x 1 ^) 1/2 ^ g = 4 x (I) (x + x 1 ^) ~ 1/2 (i + i x -i/2) + ( x + x i/2) Va (4) 

= (x +v ^)- 1/2 [2x(l + ^) + 4(x +v ^)]=(x +v ^)- 1/2 (2x + ^ + 4x + 4^) = ^l| 

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182 Chapter 3 Differentiation 



35. r: 



V cos 8 - 1 / 



dr --> / sin t 

d0 ~ Z V cos 9 - 1 / 



9 \ (cos - l)(cos 0) - (sin 9)(-sin t 



(cos 9- l) 2 



f sin 8 \ ( cos 2 3 - cos 8 + sin 2 8 \ 
Vcos - 1 ) \ (cos 9 - l) 2 J 



(2 sin 9) (1 -cost 



-2 sin 8 



(cos9-l) 3 (cosfl-1) 2 



36. r 



' sin 8 + 1 ' 



.l-cosfly " dfl ^Vl-cosfl; | (l-cos9) 2 

2(sin 9 + 1) / a „_„2 fl c ;„2 /) „;_, a\ _ 2(sin 9+ l)(cos 9-sin 9- 1) 



dr _ o /• sin 8 + \\ (1 - cos 9)(cos 0) - (sin 9 + l)(sin t 



(l-cos9) 3 



(cos — cos 2 — sirr — sin 0) 



(1 -cos9) 3 



37. y = (2x + 1) ^2x4-1 = (2x + l) 3 / 2 =}► g = § (2x + l) 1 / 2 (2) = 3^2x+l 

38. y = 20(3x - 4) 1 / 4 (3x - 4)- 1 / 5 = 20(3x - 4) 1 / 20 =>■ g = 20 (i) (3x - 4)- 19 / 20 (3) 



3 
(3x-4)i9/ 2 o 



39. y = 3 (5x 2 + sin 2x)- 3/2 => g = 3 (- |) (5x 2 + sin 2x)- 5/2 [10x + (cos 2x)(2)] = g^ffl 



40. y = (3 + cos 3 3x) 1/3 => g = - i (3 + cos 3 3x) 4/3 (3 cos 2 3x) (-sin 3x)(3) = ^J™ 2 3 * fj% 

[j ~j~ COS jXI 



41. xy + 2x + 3y = 1 => (xy' + y) + 2 + 3y' = => xy' + 3y' = -2 - y => y'(x + 3) = -2 - y => y' = - |±| 



42. x 2 +xy + y 2 -5x = 2 => 2x + (x g + y) + 2y g - 5 = =* xg + 2yg = 5-2x-y =* g (x + 2y) 

= 5-2x-y^ g = ^ 

43. x 3 + 4xy - 3y 4 / 3 = 2x => 3x 2 + (4x g + 4y) - 4y x / 3 g = 2 =► 4x | - 4y 1 / 3 | = 2 - 3x 2 - 4y 

=> S(4x-4yV3) =2 -3x 2 -4y^ g = *^£ 

44. 5x 4 / 5 + 10y 6 / 5 = 15 => 4X- 1 / 5 + ^y 1 / 5 g = =>■ ^y 1 / 5 g = -4X- 1 / 5 =>■ g = - i x -i/5 y -i/5 = _ _1_ 

45. (xy) 1 ^ = 1 => ^(xy)- 1 ^ ( x g + y ) = => x^y-l^ g = _ x -i/2 y i/2 



/2 _^ dy 
dx 



-^y => d I = -l 



46. x 2 y 2 = 1 => x 2 (2y g) + y 2 (2x) = => 2x 2 y g = -2xy 2 => g 



x+l 



2 dy _ (x+l)(l)-(x)(D _^ dy _ 1 

^ dx 



47. y 2 - ■ K 

48. y >=(H)''-' =-y' = j^ =-4v iK 



l+x\l/2 



(x+l) 2 ' ~~^ dx 2y(x+l) 2 

3 dy _ (l-x)(l)-(l+x)(-l) dy 



(1-x) 2 



dx ~~ 2y 3 (l-x) 2 



49. p 3 + 4pq-3q 2 = 2 => 3p 2 g + 4 (p + q g) - 6q = =4- 3p 2 g + 4q g = 6q - 4p 



dp 6q — 4p 

dq ~~ 3p 2 + 4q 



| (3p 2 + 4q) = 6q - 4p 



50. q = (5p 2 + 2p)- 3/2 => 1 = - | (5p 2 + 2p)- 5/2 (lOp | + 2 g) => - f (5p 2 + 2p) 5/2 = g (10p + 2) 

dp (5p 2 + 2p)° /2 

^ dq 3(5p+l) 



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Chapter 3 Practice Exercises 183 
51. r cos 2s + sin 2 s = tt =>• r(— sin 2s)(2) + (cos 2s) (^) + 2 sin s cos s = => ^ (cos 2s) = 2r sin 2s — 2 sin s cos s 

^ dr = 2r sin 2s -sin 2s = (2r- l)(sin 2s) = (2 _ j )( &) 

as cos 2s cos 2s v ' v / 



52. 2rs - r - s + s 2 = -3 =^ 2 (r + s |) - % - 1 + 2s = => f s (2s - 1) = 1 - 2s - 2r => ^ 



dr 1 - 2s - 2r 



2s- 1 



53. (a) x 3 + y 3 = 1 => 3x 2 + 3y 2 g = => g - x , - 

d 2 y _ -2xy 2 + (2yx 2 )(-^) _ -2xy 2 - ^ _ _ 2xy 3 _ 2x 4 



x 2 _ d 2 y _ y 2 (-2x)-(-^)(2yg) 



d\- 



dx 2 
,2 _ r 2 



(b) y 2 =l-f =»2y£=£ =* 1 = ^ => d | = (yx 2 r 1 => g = -(y X 2 r 2 fy(2x) + x 2 d | 



d 2 y 
dx 2 



— 2xy — x 2 I ^2 



-2xy 2 - 1 



y^x*> 



54. (a) x 2 - y 2 = 1 => 2x - 2y g = => -2y g = -2x - ^ 



(b) 



dZ _ x . d2y = y(D-*5* 
dx y dx 2 y 2 



y- x u v 2 -^ 2 



dx y 

2 v 2 



*^ = 4 (since y 2 -x 2 =-l) 



55. (a) Let h(x) = 6f(x) - g(x) => h'(x) = 6f (x) - g'(x) => h'(l) = 6f (1) - g'(l) = 6 (§) - (-4) = 7 
(b) Let h(x) = f(x)g 2 (x) =► h'(x) = f(x) (2g(x)) g'(x) + g 2 (x)f (x) => h'(0) = 2f(0)g(0)g'(0) + g 2 (0)f (0) 
= 2(l)(l)(i)+(l) 2 (-3)=-2 



(c) Leth(x) 



f(x) 



(x)+l 

(5 + !)(!) -3 (-4) _ 5_ 
(5+1) 2 12 



h'(x) 



(g(x)+l)f'(x)-f(»)g'W 
(g(x)+l) 2 



h'(l) 



(g(l) + l)f'(l)-f(l)g'(l) 
(g(l)+l) 2 



(d) Leth(x) = f(g(x)) => h'(x) = f (g(x))g'(x) =>• h'(0) = f'(g(0))g'(0) = f'(l) (|) = (i) (|) = f 

(e) Leth(x) = g(f(x)) => h'(x) = g'(f(x))f (x) =* h'(0) = g'(f(0))f (0) = g'(l)f (0) = (-4) (-3) = 12 

(f) Let h(x) = (x + f(x)) 3 / 2 =>• h'(x) = § (x + f(x)) : / 2 (1 + f (x)) =^> h'(l) = § (1 + f(l)) 1/2 (1 + f'(l)) 

= 1(1 + 3)V 2 (1 + I) = | 

(g) Let h(x) = f(x + g(x)) =► h'(x) = f'(x + g(x)) (1 + g'(x)) =► h'(0) = f'(g(0)) (1 + g'(0)) 



f'(l)(l + i) = (|) (|) 



56. (a) Let h(x) = ^f(x) =» h'(x) = ^f'(x) + f(x) • ^ => h'(l) = >/l f (1) + f(l) • A- = ± + (-3) (i 



2\/T 



B 
10 



(b) Leth(x) = (f(x))V 2 => h'(x) = \ (f(x))~V2 (f ( x) ) => h'(0) = | (f(0))- x / 2 f'(0) = \ (9)-^ 2 (-2) = - | 

(c) Leth(x) =f (y^) => h'(x) = f (-y^) " 3^ => h'(l) = f (v^) ■ ^j = H = To 

(d) Let h(x) = f(l - 5 tan x) => h'(x) = f'(l - 5 tan x) (-5 sec 2 x) =4> h'(0) = f'(l - 5 tan 0) (-5 sec 2 0) 

= f'(l)(-5)=±(-5) = -l 

(e) Leth(x) = =JE- ^ h ' (x) = (2 + cosx)fV)-f(x)(-sinx) h , (Q) = (2 + l)f'(0)-f(0)(0) = 3(_2) = _ 2 
v y v ' 2 + cosx v y (2 + cosx)^ 7 (2+iy 9 3 

(f) Leth(x)= 10sin(f)f 2 (x) => h'(x) = 10 sin (f ) (2f(x)f (x)) + f 2 (x) (10 cos (f )) (f ) 

=> h'(l) = 10 sin (|) (2f(l)f'(l)) + f 2 (1) (10 cos (f )) (f ) = 20(-3) (i) + = -12 

57. x = t 2 + tt => ^ = 2t; y = 3 sin 2x => g = 3(cos 2x)(2) = 6 cos 2x = 6 cos (2t 2 + 2tt) = 6 cos (2t 2 ) ; thus, 
f = g-f =6cos(2t 2 )-2t => %\ =6cos(0)-0 = 

I t=0 

58. t = (u 2 + 2u) 1/3 =>• £ = 5 (u 2 + 2u)~ 2/3 (2u + 2) = | (u 2 + 2u)~ 2/3 (u + 1); s = t 2 + 5t =>■ | = 2t + 5 

= 2 (u 2 + 2u) 1/3 + 5; thus | = * • | = [2 (u 2 + 2u) 1/3 + 5] (§) (u 2 + 2u)~ 2/3 (u + 1) 



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184 Chapter 3 Differentiation 



ds I 

du I u=2 



[2 (2 2 + 2(2)) 1/3 + 5] (|) (2 2 + 2(2)r 2/3 (2 + 1) = 2 (2 - 8 1 / 3 + 5) (S" 2 / 3 ) =2(2-2 + 5) (1) = | 



59. r = 8 sin (s 



dr 
ds 



cos (s + |J ; w = sin 



(aA-2) 



COS I 



l) /8sin(s + g)-2 ;thu ^^ Ja Jr a,UN-n|s f| 2! 



; (7* 



2*/8sin (s- 



dw I 
ds I s=o 



ds dr ds 



2^8sin(s + g) 



[8cos(s+|)] 



. (^8 sin (I) -2) -8 cos (f) (cos0)(8)(^) 



2j8sin(f) 



2x/4 



V~3 



60. dh + 9 = 1 => (<9 2 + t (26> f )) + f = 



1 (0 2 + 7)~ 2/3 (20) 



and 



dr I 



I 9=1 ~~ 3 



(1+7) 



-2/3 



dr I _ dr I _ dfl I 

dt I t=0 A6 I 1=0 dt I t=0 



f (20t + 1) = -<9 2 



d0 

dt 



20t+l 



; r 



7) 2/3 ;nowt = 0and(9 2 t + 6> = 1 



(c) 2 + 7) 1/3 
1 so that 



dt I t=o. 9=1 



1 



61. y + y = 2 cos x 

-2sin(0) 



3y 2 ^ 



dx 



dy 
dx 



-2 sin x 



l(3y 2 



i) 



-2 sin x 



ill 

dx 



-2 s 



3y 2 +l 



d 2 y _ (3y 2 +l)(-2cosx)-(-2 S inx)(6y£) 
0; dx 2 



3+1 "' dx 2 (3y 2 +l) 2 

d 2 y I _ (3 + l)(-2 cos 0) - (-2 sin 0)(6-0) 

dx 2 I , - (3 + 1)2 



dy 
dx 



(0,1) 



62. x 1 / 3 + y 1 / 3 = 4 => i x-' 2 / 3 + i y- 2 / 3 S = => £ 



v 2/3 



dy 
dx 



2/3 



1 . dy _ _ 
x ' dx — x 2 / 3 



d 2 y = (* 2/3 )(-ty- 1/3 S)-(-y 2/3 )(!*- 1/3 ) 

dx 2 ( x 2/3) 2 



d 2 ^ 

dx 2 



(S.Sl 



(8 2 -' 3 ) [- |-8-V 3 -(-l)] + (8 2 / 3 ) (1-8-V 3 ) 
8"/ 3 



8 2 3 



63. f(t) = jrh - and f ( l + h ) 



2(t + h)+l 
-2 



1 1 

f(t + h) - f(t) _ 2(t+h)+i 2t+i _ 2t+l-(2t + 2h+l) 
h h ~ (2t + 2h+l)(2t+l)h 

-2 



(2t + 2h+l)(2t+l)h (2t + 2h+l)(2t+l) 

-2 
(2t+l) 2 



f( t ) = lim m + h)-nt) 
W h->0 h 



h-?U (2t + 2h+l)(2t+l) 



64. g(x) = 2x 2 + 1 and g(x + h) = 2(x + h) 2 + 1 = 2x 2 + 4xh + 2h 2 + 1 

= (2x 2 + 4xh + 2h 2 i +l)-(2x 2 +l) = 4xhi2h! = 4x + 2h ^ g , (x) = ^ 



, g(x + h) - g(x) 
^ h 

g(x + h)-g(x) 
h 



lim (4x + 2h) 
h-»0 



4x 



65. (a) 



v 




> 






.V 






-1 


\ 


l 


-l 


- 


\ 



/w = 



x z ,-lS;t<0 



(b) lim f(x) = lim x 2 = and lim f(x) = lim -x 2 = => lim f(x) = 0. Since lim f(x) = = f(0) it 
x -> 0" x -» (T x -> 0+ x -> 0+ x -> x -» 

follows that f is continuous at x = 0. 

(c) lim f ' (x) = lim (2x) = and lim f ' (x) = lim (— 2x) = =>- lim f (x) = 0. Since this limit exists, it 
x -> 0" x -> 0" x -> 0+ x -> 0+ x -> 



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Chapter 3 Practice Exercises 185 



follows that f is differentiable at x = 0. 



66. (a) 



'<*) = 



f x, -1Sx<0 
itanx, OsxSn/4 



it/4 



(b) lim f(x) = lim x = and lim f(x) = lim tan x = => lim f(x) = 0. Since lim f(x) = = f(0), it 
x^O" x^O" x ^ 0+ x^0 + x^O x^O 

follows that f is continuous at x = 0. 

(c) lim f'(x) = lim 1 = 1 and lim f'(x) = lim sec 2 x=l => lim f'(x) = 1. Since this limit exists it 

x^O" x^O" x ^ 0+ x ^ 0+ x^O 

follows that f is differentiable at x = 0. 



67. (a) 



x, 0<x<] 
2-x, 1 <x<2 



(b) lim f(x) = lim x = 1 and lim f(x) = lim (2 — x) = 1 

x->l" x->l- x^l+ x^l+ 

follows that f is continuous at x = 1. 

(c) lim_ f'(x) = lim 1 = 1 and lim f'(x) = lim -1 = -1 

x —> 1 x^l x ^ 1+ x ^ 1+ 

not exist => f is not differentiable at x = 1 . 



lim f(x) = 1. Since lim f(x) = 1 = f(l), it 

X —> 1 X — > 1 



lim_ f'(x) y^ lim f'(x), so lim f'(x) does 

x^l x ^ 1+ x — > 1 



68. (a) lim f(x) = lim sin 2x = and lim f(x) = lim mx = => lim f(x) = 0, independent of m; since 

x^O" x^CT x ^o + x^O + x^O 

f(0) = = lim f(x) it follows that f is continuous at x = for all values of m. 

x — > 

(b) lim f'(x) = lim (sin 2x)' = lim 2 cos 2x = 2 and lim f'(x) = lim (mx)' = lim m = m =£- fis 

x -> 0" x -> 0" x -> 0" x -> 0+ x -> 0+ x -> 0+ 

differentiable at x = provided that lim f'(x) = lim f'(x) =>• m = 2. 

x -> x -> 0+ 

69. y = f + 2^4 = 5 x + (2x - 4)" 1 => g| = \ - 2(2x - 4)~ 2 ; the slope of the tangent is — § => - § 

= i - 2(2x - 4y 2 => -2 = -2(2x - 4)~ 2 => 1 = ^^ => (2x - 4) 2 = 1 =>■ 4x 2 - 16x + 16 = 1 
=> 4x 2 - 16x + 15 = =4> (2x - 5)(2x - 3) = =>■ x = | or x = | => (|, |) and (§, - ±) are points on the 
curve where the slope is — | . 



70. y = x-i => | = i + ^ 



=> x = ±5 => (i 



and 



1 + i ; the slope of the tangent is 3 => 3 = 1 + ^3 =>• 2=^ 
— 5, 5) are points on the curve where the slope is 3. 



71. y = 2x 3 — 3x 2 — 12x + 20 =>• ■£ = 6x 2 — 6x — 12; the tangent is parallel to the x-axis when ^ = 

=>■ 6x 2 - 6x - 12 = => x 2 - x - 2 = => (x - 2)(x + 1) = => x = 2 or x = -1 => (2, 0) and (-1, 27) are 
points on the curve where the tangent is parallel to the x-axis. 



72. y = x 3 => 



dy 



3x 2 



dy 
dx 



12; an equation of the tangent line at (—2, —8) is y + 8 = 12(x + 2) 



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186 Chapter 3 Differentiation 

=> y= 12x + 16; x-intercept: = 12x + 16 



|, 0) ; y-intercept: y = 12(0) + 16 = 16 =» (0, 16) 



73. y = 2x 3 - 3x 2 - 12x + 20 => & = 6x 2 - 6x - 12 

(a) The tangent is perpendicular to the line y = 1 — |j when -Z 



(=to) 



24; 6x 2 - 6x - 12 = 24 



=>• x 2 - x - 2 = 4 =4> x 2 -x-6 = =4> (x- 3)(x + 2) = =S> x=-2orx = 3 => (-2, 16) and (3, 1 1) are 
points where the tangent is perpendicular to y = 1 — ^ . 

(b) The tangent is parallel to the line y = \/l - 12x when ^ = -12 => 6x 2 - 6x - 12 = -12 => x 2 - x = 
=>• x(x — 1) = =>■ x = or x = 1 =>• (0, 20) and (1, 7) are points where the tangent is parallel to 
y = \fl - 12x. 



74 v = Tsmx . 
* x 



dy x(7r cos x) — (tt sin x)(l) 

dx x 2 



mi 



dy 
dx 



tt2 



Since nij = — -i- the tangents intersect at right angles. 



1 and m 2 



dy 



dx I 7T 2 

I X-— 7T 



75. 



tan x, 



f <*<f 



dy 

dx 



sec 2 x; now the slope 



of y = — | is — i =>• the normal line is parallel to 



| when -# 

2 dx 

9 1 

,S X = 2 = 



2. Thus, sec J x 



±i 

v/2 



| and x 



for - | < x < | => 



|, — l) and (f ( l) are points 
where the normal is parallel to y = — | . 




76. y = 1 



dy 
dx 



dy 
dx 



(1.0 



the tangent at (|, l) is the line y — 1 = — (x — |) 



y = —x + | + 1; the normal at 
l = (l)(x-f) => y = x-f 



1 is 




77. y = x 2 
thus, 1 



C 



:§) 2 



*■ = 2x and y = x 

C => C=i 



dx 



1 ; the parabola is tangent to y = x when 2x = 1 



1 . 

2 - 



dy 
dx 



78- y = x 3 =► I = 3x 2 =■ ;t 

intersects y = x 3 when x 3 

=> (x - a) 2 (x + 2a) = 



= 3a 2 => the tangent line at (a, a 3 ) is y — a 3 = 3a 2 (x — a). The tangent line 
3a 2 (x - a) => (x - a) (x 2 + xa + a 2 ) = 3a 2 (x - a) => (x - a) (x 2 + xa - 2a 2 ) = 
= a or x = -2a. Now p- 1 = 3(-2a) 2 = 12a 2 = 4 (3a 2 ), so the slope at 



x = —2a is 4 times as large as the slope at (a, a 3 ) where x = a. 



3 -(-2) 



79. The line through (0,3) and (5, —2) has slope m — (| _ - 

y = -x + 3; y = ^fj => ^ = (x ~ c 1)2 , so the curve is tangent to y = -x + 3 
=> (x+l) 2 =c,x/-l. Moreover, y = ^-y intersects y = — x + 3 =^> ^ 



1 => the line through (0,3) and (5, -2) is 

^z — _i 

dx ~~ L 



(1+1)2 

x+3,x/ -1 

=> c = (x + l)(-x + 3), x ^ -1. Thus c = c =>• (x + l) 2 = (x + l)(-x + 3) => (x + l)[x + 1 - (-x + 3)] 
= 0, x ^ -1 => (x + l)(2x - 2) = =>■ x = 1 (since x ^ -1) => c = 4. 



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Chapter 3 Practice Exercises 187 



80. Let ( b, ± v a 2 — b 2 j be a point on the circle x 2 + y 2 = a 2 . Then x 



dy 
dx 



±\/a 2 -b 2 



normal line through (b, ± \J a 2 — b 2 J has slope 



- 2 ^2x + 2yg=0=> g 
± v' d - h _^ normal line is 



y - ( ± \/a 2 - b 2 ) = ± v / ^ (x - b) =» y T \/a 2 - b 2 = ± v / ^ x T \A 2 - b 2 => y = ± ^^ x 
which passes through the origin. 



$1. x 2 + 2y 2 = 9 =>• 2x + 4y ^ =- () 



dx 



dy 
dx 



2y 



dy 
dx 



| and the normal line is y = 2 + 4(x — 1) = 4x — 2. 



the tangent line isy = 2— j(x— 1) 



82. x 3 + y 2 = 2 => 3x 2 + 2y ^ = 



dy _ -3x : 
dx 2y 



dy 
dx 



-fx- 



(1,1) 
2 A, ix _ 2„ 



and the normal line is y = 1 + 4 (x — 1) 



the tangent line isy= 1 + ^ (x — 1) 



83. xy + 2x - 5y = 2 =* (x | + y) + 2 - 5 g = =*■ g (x - 5) = -y - 2 =>■ 



dy 
dx 



-y-2 

x-5 



dy 
dx 



(3,2) 



the tangent line is y = 2 + 2(x — 3) = 2x — 4 and the normal line isy = 2+ -J- (x — 3) = — lx+ I . 



84. (y - x) 2 = 2x + 4 => 2(y - x) (g - l) = 2 => (y - x) g = 1 + (y - x) =► 



dy 1 + y — x 

dx y — x 



dy 
dx 



(6,2) 



the tangent line is y = 2 + | (x — 6) = |x- | and the normal line is y = 2 — I (x — 6) = — | x + 10. 



85. x 



xy = 6 



1 



*?(«£+*) 



o 



dy 



+ y = -2,/xy =>• 



dy _ -2x/xy-y 



dy 
dx 



the tangent line isy^ 1 — |(x — 4) = — |x + 6 and the normal line isy=l+^(x — 4)=^x— ^. 



86. x 3 / 2 + 2y 3 / 2 = 17 => § x 1 / 2 + 3y J / 2 g = 



dy -xV2 

dx 2yV2 



dy 
dx 



the tangent line is 



y = 4-l(x-l) 



■^ and the normal line is y = 4 + 4(x — 1) = 4x. 



87. x 3 y 3 + y 2 = x + y => [x 3 ( 3 y 2 g) + y 3 (3x 2 )] + 2y g = 1 + g => 3x 3 y 2 g + 2y g - g = 1 - 3x 2 y 3 



- ^vV + 2>-l) = l-AV - | = 3^f^T 



•-- ; - ^ ■ '-■ ix ' ? ' - S =-|,but£ is undefined. 

l(i,D ax l(i-D 

Therefore, the curve has slope — | at (1, 1) but the slope is undefined at (1, — 1). 



y = sin (x — sin x) =>• -^ = [cos (x — sin x)](l — cos x); y = =>• sin (x — sin x) = =4> x — sin x = kn, 
k = —2, —1,0, 1,2 (for our interval) => cos (x — sin x) = cos (k7r) = ± 1. Therefore, -Z = and y = when 
1 — cos x = and x = k7r. For — 2n < x < 27r, these equations hold when k = —2, 0, and 2 (since 
cos (—7r) = cos 7r = —1). Thus the curve has horizontal tangents at the x-axis for the x-values — 2ir, 0, and 2ir 
(which are even integer multiples of n) =>• the curve has an infinite number of horizontal tangents. 



x = = tan t, y = = sec t =► £ 



dy _ dy/dt _ \ sec t tan t _ tan t 



dx dx/dt 



\ sec 2 1 



sin t 



dy 
dx 



sin 



vr _ V3. t _ 



t=x/3 





=>■ x = \ tan | = 




= 2cos 3 (f)=I 


90. 


x = l+i,y = l 



f andy=Isecf = l => y = f x + \ ; g = ^ = ^ 



2cos 3 t - ^ 



dx 2 I i =7r /3 



dy 
dx 



dy/dt _ (?) 
dx/dt ~~ I 2 



dv 
dx 



(2) = -3; t = 2 => x = 1 



22 



and 



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188 Chapter 3 Differentiation 



y = l- 



-3x 



13 . _y _ dyTdt _ (- f) 
4 ' dx 2 dx/dt / 2\ 



l* 



^ - ^ (2) 3 = 6 



91. B = graph of f, A = graph of f. Curve B cannot be the derivative of A because A has only negative slopes 
while some of B's values are positive. 

92. A = graph of f, B = graph of f '. Curve A cannot be the derivative of B because B has only negative slopes 
while A has positive values for x > 0. 



93. 



(-1, 2) 




94. 



H.0) 



y = f(x) 



(t.1) 

ZX 



(6.-1) 



(1.-2) 



95. (a) 0,0 



(b) largest 1700, smallest about 1400 



96. rabbits/day and foxes/day 



97. lim 

x^0 



2x2 -X 



lim 

x^ 



' sin x \ 1 



x ) (2x-l) 



(D(i)=-1 



98. lim 

x^0 



3x — tan 7x 



lim 

x^0 



(|5 - ^V) = \ - lim (-Kr ■ s^5 . i) = 3(1.1.7) 

V2x 2xcos7x/ 2 x — * \ cos ^ x 7x (I) / * ^ 



99. lim 

r-»0 



tan 2r 



._?„ (^ • SHI? - I) = (I) CD l™ fipT = (5) (D (l) = 5 



r^0 



b (¥) 



0^0 



lim 

-> sm fl 



0^0 

sin (sin 8) .._, sin x 



/ sin (sin g) \ / sin g \ 
V_ sinfl ) \ 6 I 



100. lim 5»(™9) = lim [m^l\ (»infl) = i im S^E> . L etx = sinfl. Then x -» as 



->0 



lim 

x^0 



(9^0 



101. lim 4tan^ + t^ +1 

fl i % \ - tan 2 9 + 5 



lim 



4+-L- 



U + : 



an 2 8 J _ (4 + + 0) 
\ (1+0) 



102. lim 
0^0+ 



i - 2 cot 2 e 

5 cot 2 8 - 1 cot 6 - 8 



lim 



0+ 5--J, 



3|2fl7 



(0-2) 
(5-0-0) 



103. lim *° in * = lim xslnx , = lim xsm 2 x xU = lim 4^ • ^ 

x _, 2-2cosx x ^ 2(1- cos x) x ^ 2(2sin 2 (|)) x ^ [sin 2 (|) x 



lim 



X^ L Sln \2) S1 "l 



(1)(1)(1) = 1 



104. lim l=f^ 
0^0 " 



lim 

0^0 



2sin 2 (S) 



lim 



o L (I) 



P-^f - J] = dXl) (J) = J 



105. lim ^ 

x^O x x^0 



lim (^- • ^) = 1; let 6 = tan x => 9 -> as x -> =4> lim g(x) = lim ta " (tanx) 
> n v cos xx/ x > o x > x 



lim ^§-2 = 1. Therefore, to make g continuous at the origin, define g(0) = 1. 

0^0 e 



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106. lim f(x) 

x — > 



lim 



tan (tan x) 



q sin (sin x) 

#105); let 6 = sin x =4> 6 



lim 

x^O 

as x 



tan (tan x) 
tan x 

-> = 



continuous at the origin, define f(0) = 1 . 



sin (sin x) cos x 

■ lim -^ys 

x — » sin ' sin x ) 



1 • lim 

X — i 

lim 



Chapter 3 Practice Exercises 189 



(using the result of 



n sin (sin x) 

^3 = 1. Therefore, to make f 



■ (a) 


S = 27rr 2 + 27rrh and h constant =>■ ^ = Airr | + 27rh 


(b) 


S = 27rr 2 + 27rrh and r constant => ^ = 27rr ^ 

at at 


(c) 


S = 2m 2 + 27rrh => f = 4tit g + 2tt (r f +h gj) = 


(d) 


S constant => f = => = (4?rr + 2?rh) g + 2?rr f 



(47TT + 27Th) g 



(47rr + 27rh)£+27rr£ 



(2r + h) 



dt 

dr 
dt 



dr _ -r dh 
dt 2r+h dt 



108. S = TnVr 2 + h 2 
(a) h constant = 



. dS 
^ dt 



(b) r constant =4> g = 



ttt • (r f 2 +h | ) + Tr^r 2 + h 2 

Vr 2 + h 2 



dS 



dt Vr 2 +h 2 

S _ 7rrh dh 
It ~~ Vr 2 + h 2 dt 



% + Trx/r 2 + h 2 g = [tt ^r 2 + h 2 + 



\/r 2 + h 2 J dt 



(c) In general, f = lir^/r 2 + h 2 + -T^jf—l g + -rf^ § 

W 6 ' dt [V y f2 + h 2 J dt y r 2 + h 2 dt 

109. A = tit 2 => ^ = 27rr | ; so r = 10 and | = - f m/sec => ^ = (2tt)(10) (- f ) = -40 m 2 /sec 



110. V = s 3 



dV 
dt 



3s 



2 ds 



l = 3?f; sos = 2 0and 



dV 
dt 



1200 cm 3 /min 



ds 

dt ~~ 3(20) 2 



h* (1200) = 1 cm/min 



111. SEl 

dt 



-1 ohm/sec, ^ = 0.5 ohm/sec; and ^ = ^ + ^ =* pf = Ff-sf' Also ' 



R 



Ri 



Ri = 75 ohms and R2 = 50 ohms 



fe(-D- (55? (0.5) 



-1 dR 

(30) 2 dt 



(75} 

0.02 ohm/sec 



,5625 



- + - 
75 T 50 

500oJ = 



R = 30 ohms. Therefore, from the derivative equation, 
'- = (-900) ( 500Q ~ 5625 ^ — 9(625) 



5625-5000 . 



50(5625) 



J_ 
50 



112. <§ = 3 ohms/sec and <f = -2 ohms/sec; Z = ^R 2 + X 2 =>• f 



R I v dX 

f, d ; so that R 

^R 2 + X2 



10 ohms and 



X = 20 ohms 



dZ 
dt 



(10)(3)+(20)(-2) 
V / 10 2 + 2Q2 



—0.45 ohm/sec. 



113. Given ^7 = 10 m/sec and -# = 5 m/sec, let D be the distance from the origin =4> D 2 



x 2 + y 2 



2D 



dD 



2x l + 2 y|^ D f = x l + y|' when ( x >y) = ( 3 ' - 4 )> D = ^ 2 + (-^f = 5 



and 



(5)(10) + (12)(5) 



dD 

dt 



110 

5 



22. Therefore, the particle is moving away from the origin at 22 m/sec 



(because the distance D is increasing). 



1 14. Let D be the distance from the origin. We are given that 



dD 

dt 



1 1 units/sec. Then D 



„sm 



2D ^ 

^ dt 



,2 dx 



ZX d( -f JX dt 



x(2 + 3x) f ; x = 3 => D = ^3 2 + 3 3 



and substitution in the derivative equation gives (2)(6)(1 1) = (3)(2 + 9) ^ 



% = 4 units/sec. 

dt 



115. (a) From the diagram we have 
(b) V = i 7rr 2 h 



1 nr I 1 

I f l5 



h)^ 



47Th 3 

75 



dV 
dt 



|h. 

_ 47rh 2 dh dV 

— 25 dt ' SO dt 



-5 and h = 6 



dh 
dt 



ma ft/min - 



116. From the sketch in the text, s = rd 



ds 
dt 



dt 



dr 
dt 



Also r 



|=rf = (1.2) f. Th e refore,| 



6 ft/sec andr = 1.2 ft 



1 .2 is constant => g 
=> f = 5 rad/sec 



Copyright (c) 2006 Pearson Education 




190 Chapter 3 Differentiation 
117. (a) From the sketch in the text, ^ 



-0.6 rad/sec and x = tan 0. Also x = tan 8 



f = sec 2 6»f ;at 



point A, x = => 6 = => ^ = (sec 2 0) (-0.6) = -0.6. Therefore the speed of the light is 0.6 = | km/sec 
when it reaches point A. 

(b) ^^ • ^_ . §0sec = 18 revs/min 

v ' sec Z7r rad mm tt 



118. From the figure, a = b => i = b — 

& ' r BC r ^(,2 — r 2 

that r is constant. Differentiation gives, 



We are given 



1 da 
r " dt ~~ 

b = 2r and 



db 



b 2 -r 2 

-0.3r 



Then, 



da 



v ,:„---^(-a3r)-(2r)[-^ 



(2r) 2 - r 2 




3r 2 (-0.3r)+^a 



PI _ (3r 2 )(-0.3r)+(4r 2 )(0.3r) _ 0.3r 
3^ 



3v/3 lOv^ 



m/sec. Since $ is positive, 



the distance OA is increasing when OB = 2r, and B is moving toward O at the rate of 0.3r m/sec. 



1 19. (a) If f(x) = tan x and x = — \ , then f'(x) = sec 2 x, 



-1 andf i 



1 V 4 1 * """ * V 4 

f(x) is L(x) = 2 (x + r i) + (-1) = 2x 



2. The linearization of 



t-2 
2 



(b) If f(x) = sec x and x = — | , then f '(x) = sec x tan x, 
f (- |) = v^andf (- |) =-\fl. The linearization 
of f(x) is L(x) = -a/2 (x + f ) + a/2 



\/2(4-7r) 
4 




-tt/2 -it/4 . 
j. = -A&: + V2(4-lr|/4 



120.f(x) 



l 



1 + tan x 



f'(x) 



(1 +tanx) : 



. The linearization at x = is L(x) = f '(0)(x - 0) + f(0) = 1 - x. 



121.f(x) = a/x~+T + sin x - 0.5 = (x + l) 1 / 2 + sin x - 0.5 =>• f'(x) = (i) (x + l)" 1 / 2 + cos x 

=> L(x) = f'(0)(x - 0) + f(0) = 1.5(x - 0) + 0.5 =>■ L(x) = 1.5x + 0.5, the linearization of f(x). 



122.f(x) = £ + a/1+x - 3.1 = 2(1 - x)- 1 + (1 + x) 1 / 2 - 3.1 => f'(x) = -2(1 - x)- 2 (-l) + \ (1 + x)" 1 / 2 

- 2 | 1 



(i-x) 2 "•" 2 yr 



+ x 



L(x) = f'(0)(x - 0) + f(0) = 2.5x - 0.1, the linearization of f(x). 



123. S = 7T rx/r 2 + h 2 , r 



,2A-l/2 



constant => dS = 7r r ■ ^(r 2 + h 2 ) ' 2h dh = ,,, , dh. Height changes from h to h + dh 



dS 



Trrho(dh) 
^Th 2 



Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 



Chapter 3 Additional and Advanced Exercises 191 



12r 2 



124. (a) S = 6r 2 => dS = 12rdr. We want |dS| < (2%)S => |12rdr| < ±gl =>• |dr| < ^ . The measurement of the 
edge r must have an error less than 1%. 
(b) When V = r 3 , then dV = 3r 2 dr. The accuracy of the volume is (^) (100%) = (^f*) (100%) 



;f) (dr)(100%) 



. 100. 



(100%) = 3% 



125. C = 27rr => r = f , S = 47rr 2 = ^ , and V 



4 —3 _ Ci 

3 7rr _ 6?r2 



It also follows that dr = ^ dC, dS = — dC and 



dV = A dC. Recall that C = 10 cm and dC = 0.4 cm. 

2tH 



(a) dr=^ = ^cm 

(b) dS = f (0.4) = | cm =: 

(c) dV = (0.4) = | cm 



;f ) (100%) = (^) (^) (100%) = (.04)(100%) = 4% 



k S . 



It J V 10 > 

(100%) = (I) (^) (100%) = 8% 



'dV 

v V 



doo%) = (i)(^)(ioo%) 



12% 



126.Similar triangles yield ^ = T ^ h = 14 ft. The same triangles imply that ^±-2 = § 

120 A t 120 \ / , 1 



dh= -120a" 2 da 



da 



±£J 



h = 120a- 1 + 6 
if ) ( ± i) = ± 1 «rf ± .0444 ft = ± 0.53 inches. 



CHAPTER 3 ADDITIONAL AND ADVANCED EXERCISES 

1. (a) sin 20 = 2 sin cos =4> ^ (sin 20) = ^ (2 sin cos 0) => 2 cos 20 = 2[(sin 0)(-sin 0) + (cos 0)(cos 0)] 

=> cos 20 = cos 2 — sin 2 
(b) cos 20 = cos 2 - sin 2 =>• j- g (cos 20) = ^ (cos 2 - sin 2 0) => -2 sin 20 = (2 cos 0)(-sin 0) - (2 sin 0)(cos 0) 
=> sin 20 = cos sin + sin cos =4> sin 20 = 2 sin cos 

2. The derivative of sin (x + a) = sin x cos a + cos x sin a with respect to x is 

cos (x + a) = cos x cos a — sin x sin a, which is also an identity. This principle does not apply to the 

equation x 2 — 2x — 8 = 0, since x 2 — 2x — 8 = is not an identity: it holds for 2 values of x (— 2 and 4), but not 

for all x. 



3. (a) f(x) = cos x 



f'(x) 



-sin x =>• f"(x) = —cos x, and g(x) = a + bx + ex 2 =>■ g'(x) = b + 2cx = 
a = 1; f'(0) = g'(0) => -sin(0) = b => b = 0; f"(0) = g"(0) 



g"(x) = 2c; 



Therefore, g(x) = 1 — \ x 2 . 



'(x) = b cos x — c sin x; also, 
cos (a) = b cos (0) — c sin (0) 



also, f(0) = g(0) => cos (0) = a 
=> -cos(0) = 2c => c = - \, 

(b) f(x) = sin (x + a) => f '(x) = cos (x + a), and g(x) = b sin x + c cos x = 
f(0) = g(0) => sin (a) = b sin (0) + c cos (0) => C = sin a; f '(0) = g'(0) 

=> b = cos a. Therefore, g(x) = sin x cos a + cos x sin a. 

(c) When f(x) = cos x, f'"(x) = sin x and f ' 4 '(x) = cos x; when g(x) = 1 — I x 2 , g'"(x) = and g' 4 '(x) = 
Thus f '"(0) = = g'"(0) so the third derivatives agree at x = 0. However, the fourth derivatives do not 
agree since f ™(0) = 1 but g' 4 '(0) = 0. In case (b), when f(x) = sin (x + a) and g(x) 

= sin x cos a + cos x sin a, notice that f(x) = g(x) for all x, not just x = 0. Since this is an identity, we 
have f * n '(x) = g^°'( x ) f° r an Y x an d any positive integer n. 



= —sin x => y + y = 
-cos x + cos x = 0; y 



—sin x + sin x = 0; y = cos x =4> y' = —sin x 
a cos x + b sin x => y' = — a sin x + b cos x 



(a) y = sin x =4> y = cos x =>■ y 
=>• y" = —cos x => y" + y = - 

=>• y" = —a cos x — b sin x => y" + y = (—a cos x — b sin x) + (a cos x + b sin x) = 

(b) y = sin(2x) => y' = 2cos(2x) => y" = -4sin(2x) =>■ y" + 4y = -4 sin(2x) + 4 sin(2x) = 0. Similarly, 
y = cos (2x) and y = a cos (2x) + b sin (2x) satisfy the differential equation y' + 4y = 0. In general, 

y = cos (mx), y = sin (mx) and y = a cos (mx) + b sin (mx) satisfy the differential equation y" + m 2 y = 0. 



Copyright (c) 2006 Pearson Education 




192 Chapter 3 Differentiation 



5. If the circle (x — h) 2 + (y — k) 2 = a 2 and y = x 2 + 1 are tangent at (1, 2), then the slope of this tangent is 



= 2x| 
2x 



2 and the tangent line is y = 2x. The line containing (h, k) and (1, 2) is perpendicular to 



k-2 
h- 1 



h = 5 — 2k => the location of the center is (5 — 2k, k). Also, (x — h) 2 



(y - k) 2 = a 2 



=» x - h + (y - k)y' = 0^1 + (y') 2 + (y - k)y" 







y 



n _ i + (yT 



k-y 



At the point (1, 2) we know 



y' = 2 from the tangent line and that y" = 2 from the parabola. Since the second derivatives are equal at (1, 2) 



we obtain 2 



1 + (2) 2 



k-2 "^ iv 2 

lies on the circle we have that a 



. Then h 

~~ 2 ■ 



2k 



the circle is (x + 4) 2 + (y 



I) 2 



a 2 . Since (1,2) 



6. The total revenue is the number of people times the price of the fare: r(x) = xp = x (3 — J| 



40 ) 



< x < 60. The marginal revenue is j^ = (3 



2x 3 



dx ~~ V-' 40 
40M* 40 V ' 1 "^" dx 

are on the bus the marginal revenue is zero and the fare is p(40) 



40 ) 



40 J 



dr 
dx 



40 



where 
)[(3- 



40 ) 



2x1 
40 J 



3 (3 ^)(1 tti) ■ Then * = =4> x = 40 (since x = 120 does not belong to the domain). When 40 people 



40) 



$4.00. 



7. (a) y = uv => ^ = ^ v + u ^ = (0.04u)v + u(0.05v) = 0.09uv = 0.09y =>- the rate of growth of the total production is 



9% per year 
(b) If 

year 



dy 



f = -0.02u and g = 0.03v, then ^ = (-0.02u)v + (0.03v)u = O.Oluv = O.Oly, increasing at 1% per 



8. When x 2 + y 2 = 225, then y' = - £ . The tangent 
line to the balloon at (12, -9) is y + 9 = f (x - 12) 
=> y = | x — 25. The top of the gondola is 15 + 8 
= 23 ft below the center of the balloon. The inter- 
section of y = —23 and y = | x — 25 is at the far 



right edge of the gondola 



-23 



x-25 



Thus the gondola is 2x = 3 ft wide. 



x 2 + y 2 = 225 




(-12,-9) 



\ 

>, 
Suspension cables — \- 

Gondola 



y-(4/3)x-25 8ft 

L 



-Width 



NOT TO SCALE 



9. Answers will vary. Here is one possibility. 



L 



10. s(t) = 10 cos (t 



v(t) 



ds 
dt 



10 cos 



10 



(a) s(0)- KAl 

(b) Left: -10, Right: 10 

(c) Solving 10 cos (t+ |) = -10 
Solving 10 cos (t+ I) = 10 = 
is farthest to the right. Thus, v 



(d) Solving 10 cos (t 







10 sin t 



=^ cos (t 

- COS (t + 
3tt> 



a(t) 



dv 
dt 



3-a 



d£s 
dt- 1 



10 cos (t 



— 1 => t = t when the particle is farthest to the left. 



1 



0, v 



'llT\ 



0, a 



10, Ivi 



, but t > => t = 2-k - 
10, and a ( 7 f) = -10. 
)\ = lOandaf?) = 0. 



— - 

4 



when the particle 



Copyright (c) 2006 Pearson Education 




Chapter 3 Additional and Advanced Exercises 193 



11. (a) s(t) = 64t - 16t 2 =>• v(t) = f t = 64 - 32t = 32(2 - t). The maximum height is reached when v(t) = 

=>• t = 2 sec. The velocity when it leaves the hand is v(0) = 64 ft/sec. 
(b) s(t) = 64t - 2.6t 2 =>• v(t) = ^ = 64 - 5.2t. The maximum height is reached when v(t) = => t w 12.31 sec. 
The maximum height is about s(12.31) = 393.85 ft. 

12. si = 3t 3 - 12t 2 + 18t + 5 and s 2 = -t 3 + 9t 2 - 12t =4> v : = 9t 2 - 24t + 18 and v 2 = -3t 2 + 18t - 12; v x = v 2 

=^ 9t 2 - 24t + 18 = -3t 2 + 18t - 12 => 2t 2 - 7t + 5 = => (t - l)(2t - 5) = => t = 1 sec and t = 2.5 sec. 

13. m(v 2 -v 2 )=k(x 2 -x 2 ) => m (2vf)=k(-2xf) => m| = k(-|) | => m £ = -kx (i) f . Then 
substituting $ = v =>• m ^ = — kx, as claimed. 



14. (a) x = At 2 + Bt + C on [ti , t 2 ] 



2At + B =>• v (^ j2 ) = 2A (^y^ 2 ) + B = A (ti + t 2 ) + B is the 



Ax 
At 



instantaneous velocity at the midpoint. The average velocity over the time interval is v a , 

_ (At| + Bt 2 + C)-(Atf + B tl + C) _ (t 2 - tl )[A(t 2 +ti)+B] = A U , j. \ , g 
fe — ti t 2 — ti ^ ' 

(b) On the graph of the parabola x = At 2 + Bt + C, the slope of the curve at the midpoint of the interval 
[ti , t 2 ] is the same as the average slope of the curve over the interval. 



15. (a) To be continuous at x = 7r requires that lim_ sin x = lim (mx + b) => = van + b =>• m = — 



(b) 



if y' = { 



COS X X "-C 7T 

' _ is differentiable at x = 7r, then lim_ cos x = m => m = — 1 and b = n. 

m, X > 7T X -> 7T 



16. f(x) is continuous at because lim 



= f(0). f'(0) 



lim 



f(x) - f(0) 



lim 



iiL_0 



lim 



1 — COS X \ / 1 + COS X > 
X 2 / V 1 + COS X I 



lim (^ 

x^O V x ; 



, 1 + COS X ) 



x -° x-»"0 

\ . Therefore f '(0) exists with value \ 



lim_ f(x) = f(2) 



f'(x) = | 



In order that f '(2) exist we must have a = 2a(2) — b =^> a = 4a — b 



3a 



a = | and b 



17. (a) For all a, b and for all x ^ 2, f is differentiable at x. Next, f differentiable at x = 2 =>• f continuous at x = 2 

2a = 4a - 2b + 3 => 2a - 2b + 3 = 0. Also, f differentiable at x ^ 2 

a, x < 2 
2ax - b, x > 2 

Then 2a — 2b + 3 = and 3a = b 

t t 

(b) For x < 2, the graph of f is a straight line having a slope of § and passing through the origin; for x > 2, the graph of f 
is a parabola. At x = 2, the value of the y-coordinate on the parabola is | which matches the y-coordinate of the point 
on the straight line at x = 2. In addition, the slope of the parabola at the match up point is | which is equal to the 
slope of the straight line. Therefore, since the graph is differentiable at the match up point, the graph is smooth there. 

18. (a) For any a, b and for any x/ — 1, g is differentiable at x. Next, g differentiable at x = — 1 => g continuous at 

x = — 1 =>- lim g(x) = g(— 1) =^ -a-l+2b = -a + b =>• b = 1. Also, g differentiable at x / - 1 

X — » — 1 + 



I'M = { 



a, x < -1 
3ax 2 + 1, x > -1 



In order that g'(— 1) exist we must have a = 3a(— l) 2 + 1 =S> a=3a+l 



(b) For x < — 1, the graph of f is a straight line having a slope of — \ and a y-intercept of 1 . For x > — 1 , the graph of f is 
a parabola. At x = — 1 , the value of the y-coordinate on the parabola is | which matches the y-coordinate of the point 
on the straight line at x = — 1. In addition, the slope of the parabola at the match up point is — | which is equal to the 
slope of the straight line. Therefore, since the graph is differentiable at the match up point, the graph is smooth there. 



19. fodd 



f(-x) 



-f(x) 



d (f(-x)) = -f (-f(x)) =^ f'(-x)(-l) = -f'(x) => f'(-x) = f'(x) =>■ f is even. 



dx 



Copfiigl (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 



194 Chapter 3 Differentiation 



20. feven =► f(-x) = f(x) =► A (f(- x )) = A (f(x)) 



f'(-x)(-l) = f'(x) => f (-x) = -f (x) 



f is odd. 



21. Leth(x) = (fg)(x) = f(x)g(x) => h'(x) = lim 

A > AQ 



h(x) - h(x ) _ j^ m 
x - x x — > x 



f(x)g(x)-f(x )g(x ) 

X-XQ 



lim 

x^ x 



-x^xo [fwl^f 21 ]] +x 1 iPl h [ ^ il 



f(x) g(x) - f(x) g(x ) + f(x) g(x ) - f(x ) g(x ) 



f(xo) x lim o «[=*U + g(x )f (xo) = ■ x lim Q 



x-*o J x —> x L DX "' |_ x ~ x o I 

^5^1 + g(x ) f (xo) = g(x ) f '(Xq), if g is 



continuous at Xq. Therefore (fg)(x) is differentiable at Xq if f(xg) = 0, and (fg)' (xq) = g(xg) f '(xq). 



22. From Exercise 21 we have that fg is differentiable at if f is differentiable at 0, f(0) = and g is continuous 
atO. 

(a) If f(x) = sin x and g(x) = |x| , then |x| sin x is differentiable because f'(0) = cos (0) = 1, f(0) = sin (0) = 
and g(x) = |x| is continuous at x = 0. 

(b) If f(x) = sin x and g(x) = x 2 / 3 , then x 2 / 3 sin x is differentiable because f'(0) = cos (0) = 1, f(0) = sin (0) = 
and g(x) = x 2 / 3 is continuous at x = 0. 



(c) If f(x) = 1 — cos x and g(x) 



/ x, then \/x(l — cos x) is differentiable because f'(0) = sin (0) = 0, 



f(0) = 1 cos (0) = and g(x) = x 1//3 is continuous at x = 0. 
(d) If f(x) = x and g(x) = x sin Q) , then x 2 sin Q) is differentiable because f'(0) = 1, f(0) = and 

(so g is continuous at x = 0). 



lim x sin (-) 

x^0 Vx; 



lim 

x^0 



lim ^ 

t — ► oo ' 



23. If f(x) = x and g(x) = x sin (-) , then x 2 sin (-) is differentiable at x = because f'(0) = 1, f(0) = and 



lim x sin (-^ 

x^0 Vx/ 



lim 



ffi 



lim 

t — » 00 



(so g is continuous at x = 0). In fact, from Exercise 21, 



h'(0) = g(0)f'(0) = 0. However, forx ^ 0, h'(x) = [x 2 cos (1)] (- ^) + 2x sin Q) . But 
lim h'(x) = lim [—cos (-) + 2x sin (i)l does not exist because cos (-) has no limit as x — > 0. Therefore, 

x^O x^O L Vx7 Vx;J VxV 

the derivative is not continuous at x = because it has no limit there. 
24. From the given conditions we have f(x + h) = f(x) f(h), f(h) 1 = hg(h) and lim g(h) = 1. Therefore, 

h — > 



f'(x) = lim 



f(x+h) - f(x) 



lim 



f(x)f(h)-f(x) 



lim f(x) [Sflpi 
h -► L h 



f(x) lim g(h) 

Ln — > u 



f(x) • 1 = f(x) 



f'(x) = f(x) and f'(x)exists at every value of x. 



dy 



25. Step 1: The formula holds for n = 2 (a single product) since y = U1U2 - , 
Step 2: Assume the formula holds for n = k: 

dv dui 1 dn~ 

y = uiu 2 ---u k => ai = dF u 2 u 3 ---u k + 
Ify = uiu 2 ---u k u k+ i = (u 1 u 2 ---u k )u k+1 ,then ^ = ; iv ' u k _ 



£ U 2 + Ui 



U1II2— Uti £ 



(^u 2 u 3 ---u k + Ui ^ u 3 ---u k + ••• 

^U 2 U 3 ---U k+1 +Ul ^U 3 ---U k+1 + 



Hi], ^ 

UiU 2 ---U k _, -j±) lit 



dx 
din 
dx , 



UlU 2 ---U k ^±l 



u lU2 -u k ^ 



UiU 2 ---U k 



du. 



+ U!U 2 ---U! 



du,. 



Thus the original formula holds for n = (k+1) whenever it holds for n = k. 



26. Recall 



„ ™' , ■ Then(™ 

k! (m — k)! VI. 



l!(m-l)! 



m and 



m!(k+l) + m!(m-k) 



m!(m+ 1) 



(m+1)! 



(k+l)!(m-k)! (k+l)!(m-k)! (k+ 1)! ((m + 1) - (k+ 1))! 

Leibniz's rule by mathematical induction. 
Stepl: Ifn=l,then^ 



Vk+W k!(m-k)! 

[k+i ) ■ Now, we prove 



(k+l)!(m-k-l)! 



U 3x + v 3x ■ Assume that the statement is true for n = k, that is: 



d'-(iiv) 
dx 1 



d k u .. I u d k ] u dv i /k\ d k 2 u d 2 v i 

l V ~t~ K A„k~l Av ' \ 1 I ^»k-2 A^2 T 



Step 2: Ifn = k+1, then 



dx k dx k_1 dx 

d k+ '(uv) 



,2) dx k - 2 dx 2 

d_ 

dx 



+ 



k \ du d k ~'v 
k - l) dv dx k -' 



d*v 
dx k 



I d k (uv) \ __ r d k +'u , d"u dv] , |V d k u dv , ^ d k ~'u d 2 v ] 
\ dx k ) ~ Ldx k +! v "t" dx k dxJ + [ K dx k dx + K dx k -i dx 2 J 



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Chapter 3 Additional and Advanced Exercises 195 



k\ d k -'u d 2 v , 
.2) dx k -i dx 2 " t " 

du ft 1 d k+1 u ] 
dx dx k "r" u dx k +ij 



'k\ d k ~ 2 u d 3 v 
^2/ dx k - 2 dx 3 



' k \ ft d k ~'v 
,k-lJ dx 2 dx k -i 



\ k— 1 / dx dx 1 



du d u .. 



+ [ 

+ [( k *i) + (*)]££ + 



d^v + (k+i) 



du dv 
dx k dx 



[(l) + (!)]DS 



'k+l\ du dS; 
\ k / dx dx k 



du d^v 

dx dx k 

d k+1 v 



d ki 'v 
dx 1 * 1 



^v + (k+l)0^ 



'k+l\ d M u d 2 v 



dx 1 - 



dx-' 



+ . 



U 



dx k +! • 



Therefore the formula (c) holds for n = (k + 1) whenever it holds for n = k. 



27. (a) T 2 = ^ ^ L = ^ L = Lligg^ztt^j ^ L ^ agl5fi ft 

(b) T 2 = ^ ^ T = ^Lx/l; dT = ^L . 1 dL = -^dL; dT = ,, noir , J 1 ^^, ,, (0.01 ft) « 0.00613 sec. 

v ' g x/g x/g 2x/L x/Lg x/(0.8156ft)(32.2 ft/sec 2 ) v ' 

(c) Since there are 86,400 sec in a day, we have (0.00613 sec)(86,400 sec/day) « 529.6 sec/day, or 8.83 min/day; the 
clock will lose about 8.83 min/day. 



= s 3 
2k 



dt 

so - 



3s 



2ds 



-k(6s 2 ) 



dt 



2k. If So = the initial length of the cube's side, then Si = &o — 2k 

(vo) 1/3 



(Ml 3 



Si. Let t = the time it will take the ice cube to melt. Now, t = S 
11 hr. 



so 

S0-S1 



-(boY 



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196 Chapter 3 Differentiation 
NOTES: 



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CHAPTER 4 APPLICATIONS OF DERIVATIVES 



4.1 EXTREME VALUES OF FUNCTIONS 

1. An absolute minimum at x = C2, an absolute maximum at x = b. Theorem 1 guarantees the existence of such 
extreme values because h is continuous on [a, b]. 

2. An absolute minimum at x = b, an absolute maximum at x = c. Theorem 1 guarantees the existence of such 
extreme values because f is continuous on [a, b]. 

3. No absolute minimum. An absolute maximum at x = c. Since the function's domain is an open interval, the 
function does not satisfy the hypotheses of Theorem 1 and need not have absolute extreme values. 

4. No absolute extrema. The function is neither continuous nor defined on a closed interval, so it need not fulfill 
the conclusions of Theorem 1. 

5. An absolute minimum at x = a and an absolute maximum at x = c. Note that y = g(x) is not continuous but 
still has extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the 
hypothesis is not satisfied, absolute extrema may or may not occur. 

6. Absolute minimum at x = c and an absolute maximum at x = a. Note that y = g(x) is not continuous but still 
has absolute extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when 
the hypothesis is not satisfied, absolute extrema may or may not occur. 

7. Local minimum at ( — 1, 0), local maximum at (1, 0) 

8. Minima at (-2, 0) and (2, 0), maximum at (0, 2) 

9. Maximum at (0, 5). Note that there is no minimum since the endpoint (2, 0) is excluded from the graph. 

10. Local maximum at (—3, 0), local minimum at (2, 0), maximum at (1, 2), minimum at (0, —1) 

11. Graph (c), since this the only graph that has positive slope at c. 

12. Graph (b), since this is the only graph that represents a differentiable function at a and b and has negative slope at c. 

13. Graph (d), since this is the only graph representing a funtion that is differentiable at b but not at a. 

14. Graph (a), since this is the only graph that represents a function that is not differentiable at a or b. 



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198 Chapter 4 Applications of Derivatives 



15. f(x) = | x - 5 => f (x) 
f(-2) = - ¥, f(3) = -3 



=> no critical points; 
the absolute maximum 



19 



is —3 at x = 3 and the absolute minimum is — y at 




16. f(x) = —x — 4 =>- f (x) = — 1 => no critical points; 
f(— 4) = 0, f(l) = — 5 =>• the absolute maximum is 
at x = —4 and the absolute minimum is —5 at x = 1 




(1,-5) 



17. f(x) = x 2 — 1 =>• f'(x) = 2x =4* a critical point at 
x = 0; f(-l) = 0, f(0) = -1, f(2) = 3 => the absolute 
maximum is 3 at x = 2 and the absolute minimum is — 1 
atx = 




18. f(x) = 4 - x 2 =4> f (x) = -2x => a critical point at 
x = 0; f(-3) = -5, f(0) = 4, f(l) = 3 => the absolute 
maximum is 4 at x = and the absolute minimum is —5 
at x = — 3 



/(x) = 4-x 




19. F(x) 



F'(x) = 2x" 



, however 



x = is not a critical point since is not in the domain; 
F(0.5) = —4, F(2) = —0.25 => the absolute maximum is 
—0.25 at x = 2 and the absolute minimum is —4 at 
x = 0.5 



1 


v 









1 . 


(2,-0.25) 




-1 




Abs max 




-2 


jy = 


4;,0.5<i<2 

X 




-3 






-4 


- • (0.5, 


-4) 






Abs 


nin 





Copyright (c) 2006 Pearson Education 




Section 4. 1 Extreme Values of Functions 199 



20. F(x) 



F'(x) 



, however 



x = is not a critical point since is not in the domain; 
F(— 2) = | , F(— 1) = 1 =£- the absolute maximum is 1 at 
x = — 1 and the absolute minimum is k at x = — 2 



(-1,1) 



(-2,1/2) 



F(x) = — 



-15 -1 -0.5 



y 

i 



21. h(x) = \/x = x 1 / 3 =>■ h'(x) 



I ,,-2/3 
3 A 



a critical point 

at x = 0; h(-l) = -1, h(0) = 0, h(8) = 2 =>• the absolute 
maximum is 2 at x = 8 and the absolute minimum is — 1 
at x = - 1 



2 _ -1<.t<8 „ (8. 2) 
1 V^"""^ max 


M 
(-1,-D 

Abs min 


12 3 4 5 6 7 8 



22. h(x) = -3x 2 / 3 => h'(x) = -2x -1 / 3 => a critical point at 
x = 0; h(-l) = -3, h(0) = 0, h(l) = -3 => the absolute 
maximum is at x = and the absolute minimum is —3 
at x = 1 and at x = — 1 



(-1,-3) 



h(x) = -3* 2/3 




(1,-3) 



23. g(x) = \/4-x 2 = (4 - x 2 ) 1/2 
=> g'(x) = i (4 - x 2 )" 1/2 (-2x) 



v/4-x 2 

=>- critical points at x = —2 and x = 0, but not at x = 2 
because 2 is not in the domain; g(— 2) = 0, g(0) = 2, 

g(l) = V 3 =>■ the absolute maximum is 2 at x = and the 

absolute minimum is at x = —2 





t 


* 




y = ^4-^ 




(0, 2) Abs max 




-2<.*<1^ 


1 


- 












(-2,0) - 1 








Abs 








min 









24. g(x) = -A/5-x 2 = 



(5 



x 2 ) 1/2 (5 - x 2 )~ 1/2 (-2x) 
=> critical points at x = — y 5 



and x = 0, but not at x = y 5 because y 5 is not in the 

domain; f (- V^) = 0, f(0) = -\/5 

=>■ the absolute maximum is at x = — y 5 and the absolute 
minimum is — y 5 at x = 



-2.5 -2 -1.5 -1 .0.5 



(-/5\0) 




«0O = -V5^7 



(0,-v^) 



25. f(0) = sin i 



f (0) = cos (9 =>■ = f is a critical point, 



but = -rp is not a critical point because -f is not interior to 



the domain; f ( 



-i,f(!) = i,f($) 



=4> the absolute maximum is 1 at 6* = | and the absolute 
minimum is — 1 at = =2 



(-ir/2,-1) 
Abs min 



(tt/2, 1) Abs max 



-J— 



■nil 5w!6 
y=sm8,-ir/2<0< 5irl6 



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200 Chapter 4 Applications of Derivatives 



26. f(9) = tan 9 => f'(9) = sec 2 => f has no critical points in 
(^p, ^) . The extreme values therefore occur at the 

endpoints: f (=^) = — y3andf (|) = 1 =^ the absolute 
maximum is 1 at 6 = | and the absolute 



minimum is — y 3 at 



— 7T 

3 



ir/4,1) 




-it/3, -/T) 



27. g(x) = esc x =£- g'(x) = —(esc x)(cot x) => a critical point 



at x 



73-SV2, 



i.g(¥) 



the 



absolute maximum is -7- at x = f and x = 7? , and the 
absolute minimum is 1 at x = 5 



Abs max Abs max 

(tt/3, 2/Vl) (2tt/3, 2/VI) 



J-CSC.I (77/2,1) 

77/3<^<277/3 ^ bs 

min 



77/3 77/2 277/3 



28. g(x) = sec x => g'(x) = (sec x)(tan x) =>■ a critical point at 

the absolute 



x = 0;g(-f)=2,g(0)=l,g(f) = ^ 



maximum is 2 at x 
atx = 



I and the absolute minimum is 1 



(-tt/3,2 




(Tt/6,2/vf) 



g (jr) s sec* 



-1 -0.8 -0.6 -0.4 -0.2 0.2 0.4 



2\l/2 



29. f(t) = 2 — |t| = 2 — a/^" = 2 — (t 2 ) 

^f(t)=-i(t 2 r 1/2 (2t) = --^ 



=S> a critical point at t = 0; f(— 1) = 1, 
f(0) = 2, f(3) = 1 =>• the absolute maximum is 2 at t = 
and the absolute minimum is — 1 at t = 3 





(0, 2) Abs 




\ max 


» 1 


\ y = 2-l/l 




\ -1<(<3 


-1 


1 ZV 3 


-1 


AbsN^ 




nun (3] _i) 



30. f(t) = |t - 5 1 = ^(T^ = (ft - 5ff 2 => f ft) 



i((t-5) 2 )" 1/2 (2(t-5)) 



t-5 

|t-5| 



2 VV "> I \^ ->!> ^(1 -5)2 

=4> a critical point at t = 5; f(4) = 1, f(5) = 0, f(7) = 2 
=>■ the absolute maximum is 2 at t = 7 and the absolute 
minimum is at t = 5 




f(t) = \t-5\ 



31. f(x) = x 4 / 3 => f (x) = I x 1 / 3 =>• a critical point at x = 0; f(- 1) = 1, f(0) = 0, f(8) = 16 =>■ the absolute 
maximum is 16 at x = 8 and the absolute minimum is at x = 

32. f(x) = x 5 / 3 => f'(x)=fx 2 / 3 =>• a critical point at x = 0; f(- 1) = -1, f(0) = 0, f(8) = 32 =>■ the absolute 
maximum is 32 at x = 8 and the absolute minimum is — 1 at x = — 1 



33. g(<9) = 6> 3 / 5 => g'(6)= \9- 2 l 5 



-, a critical point at = 0; g(— 32) 
maximum is 1 at 9 = 1 and the absolute minimum is — 8 at 9 = —32 



-8, g(0) = 0, g(l) = 1 => the absolute 



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Section 4. 1 Extreme Values of Functions 201 

34. h(0) = 30 2 / 3 => h'(0) = IQ- 1 ^ => a critical point at = 0; h(-27) = 27, h(0) = 0, h(8) = 12 => the absolute 
maximum is 27 at = —27 and the absolute minimum is at 8 = 



35. Minimum value is 1 at x = 2. 



-2 



-2 




[-2,6] by [-2,4] 



36. To find the exact values, note that y' = 3x 2 — 2, 



which is zero when x = ± 



Local maximum at 



-,/§,4 



4y/C 

9 



(-0.816, 5.089); local 



minimum at 



(\/i> 4 _ ^) ~ (°- 816 ' 2 - 911 ) 




6,6] by [-2,7] 



37. To find the exact values, note that that y' = 3x 2 + 2x — 8 
= (3x — 4)(x + 2), which is zero when x = — 2 or x = |. 



Local maximum at (—2, 17); local minimum at (I, 



.in 

27 1 




2 4 6 



6,6] by [-5,20] 



38. Note that y' = 3x 2 - 6x + 3 = 3(x - l) 2 , which is zero at 
x = 1. The graph shows that the function assumes lower 
values to the left and higher values to the right of this point, 
so the function has no local or global extreme values. 




6,6] by [-4,4] 



39. Minimum value is when x 



- 1 or x = 1 . 




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202 Chapter 4 Applications of Derivatives 
40. The minimum value is 1 at x = 0. 




-l l 

[-1.5,1.5] by [-0.5,3] 



41. The actual graph of the function has asymptotes at x = ± 1, 
so there are no extrema near these values. (This is an 
example of grapher failure.) There is a local minimum at 

(0,1). 




[-4.7,4.7] by [-3.1,3.1 



42. Maximum value is 2 at x = 1; 

minimum value is at x = —1 and x = 3. 




[-4.7,4.7] by [-3.1,3.1] 



43. Maximum value is \ at x = 1; 
minimum value is — | as x = — 1. 




-5,5] by [-0.7,0.7] 



44. Maximum value is \ at x = 0; 
minimum value is — \ as x = —2. 




-5,5] by [-0.8,0.6] 



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Section 4. 1 Extreme Values of Functions 203 



45. y' = x 2 / 3 (l) + fx-^x + 2 ) = |+i 



cnt. pt. 



derivative 




undefined 



extremum 



local max 
local min 



value 



iflO 1 / 3 = 1.034 




-3 x 
[-4,4] by [-3,3] 



46. 



8x 2 -8 



y' = x 2 / 3 (2x) + |x- 1 / 3 (x 2 - 4) = s -f0 



crit. pt. 


derivative 


extremum 


value 


x= -1 
x = 
x= 1 




undefined 




minimum 
local max 
minimum 


-3 

3 



y 

\ 2 

\ 1 / 

4 ^2 i 2 4 

\ -i/\ j 

v^ 3 1 vy 



-4,4] by [-3,3] 



47. y' = x -l ( - 2x) + (1)^4 -x2 



^2^4-x 2 
-x 2 + (4-x 2 ) 



4-2x 2 



V4 — x 2 V4 — x 2 



crit. pt. 


derivative 


extremum 


value 


x= -2 


undefined 


local max 





x = -y2 





minimum 


-2 


x= s/2 





maximum 


2 


x = 2 


undefined 


local min 






h 
3 - 

2 - .. 

1 •^ \ 

"1 ' ~7 ' ' * 

-2 -1 / 1 2 

^ -^ -2 - 

-3 - 



[-2.35,2.35] by [-3.5,3.5] 



-x 2 + (4x)(3-x) _ _5x 2 + 12x 
2%/3-x " 2^/3 -x 



crit. pt. 





12 
5 
3 



derivative 







undefined 



extremum 



minimum 
local max 
minimum 



value 





144 15 l/2 
125 LO 





4.462 




[-4.7,4.7] by [-1,5] 



49. y' 



-2, x<l 
1, x> 1 



crit. pt. 


derivative 


extremum 


value 


x= 1 


undefined 


minimum 


2 




-2 2 4 

-4.7,4.7] by [0,6.2] 



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204 Chapter 4 Applications of Derivatives 

50 y'=( - 1 ' X< ° 

y 1 2 - 2x, x > 



crit. pt. 


derivative 


extremum 


value 


x = 
x= 1 


undefined 




local min 
local max 


3 

4 




[-4,4] by [-1,6] 



51. y' 



-2x - 2, 
-2x + 6, 



x< 1 
x> 1 



crit. pt. 


derivative 


extremum 


value 


x= -1 
x= 1 
x = 3 




undefined 




maximum 
local min 
maximum 


5 
1 

5 




-4,6] by [-2,6] 



f -I x 2 - I X + 11 

52. We begin by determining whether f'(x) is defined at x = 1, where f(x) = < 4 2 4 

[x 6 - 6x 2 + 8x, 



x< 1 
x> 1 



Clearly, f'(x) = -H- \ if x < 1, and lim f'(l + h) = -1. Also, f'(x) = 3x 2 - 12x + 8ifx> 1, and 



lim f ' ( 1 + h) = — 1 . Since f is continuous at x = 1 , we have that f ' ( 1 ) 

h->0+ 



-l.Thus, 



f'(x) 



3x 2 - 12x + I 



x < 1 
x> 1 



Note that -\n- \ = when x = -1, and 3x 2 - 12x + 1 



when x 



12±yi2 2 -4(3)(8) _ 12±y / 48 



But 2 



2y/3 



0.845 < 1, so the critical points occur at x = —1 and x = 2 



2(3) 
1-sfl „ 



2± 



2^3 



3.155. 



crit. pt. 


derivative 


extremum 


value 


x = -l 

x* 3.155 






local max 
local min 


4 
» -3.079 




-4,6] by [-5,5] 



53. (a) No, since f'(x) = |(x - 2) 1/3 , which is undefined at x = 2. 

(b) The derivative is defined and nonzero for all x/2. Also, f(2) = and f(x) > for all x^2. 

(c) No, f(x) need not have a global maximum because its domain is all real numbers. Any restriction of f to a closed 
interval of the form [a, b] would have both a maximum value and minimum value on the interval. 

(d) The answers are the same as (a) and (b) with 2 replaced by a. 



54. Note that f( 



»> = { 



-x 3 + 9x, x< -3or0 <x<3 ^ c rl , . f -3x 3 + 9, x<-3or0<x<3 

, . Therefore, fix) = < ., 

x 3 -9x, -3<x<0orx>3 w \ 3x 3 - 9, -3<x<0orx>3 

(a) No, since the left- and right-hand derivatives at x = 0, are —9 and 9, respectively. 

(b) No, since the left- and right-hand derivatives at x = 3, are —18 and 18, respectively. 



Copyright (c) 2006 Pearson Education 




Section 4. 1 Extreme Values of Functions 205 



(c) No, since the left- and right-hand derivatives at x = —3, are 18 and —18, respectively. 

(d) The critical points occur when f (x) = (at x = ± \J 3) and when f (x) is undefined (at x = and x = ±3). The 
minimum value is at x = —3, at x = 0, and at x = 3; local maxima occur at ( — v 3, 6\/3) and ( \/3, 6v3). 



55. 



DF 






-H^H 



•R 



x i 9-x 

(a) The construction cost is C(x) 
a graph of C(x). 



O.Z^/AS 



0.2(9 — x) million dollars, where < x < 9 miles. The following is 



^^ 


2.95 


J2 

M 
1 


2.9 
2.85 


£ 

o 


2.8 
2.75 

2.7 
2.65 




8 



9 



Solving C'(x) 



0.3x 
V16 + X2 



0.2 = gives x = ± 



8\A 



4 5 6 
x (miles) 

w ±3.58 miles, but only x = 3.58 miles is a critical point is 



the specified domain. Evaluating the costs at the critical and endpoints gives C(0) = $3 million, CI -^- J « $2,694 

million, and C(9) « $2,955 million. Therefore, to minimize the cost of construction, the pipeline should be placed 
from the docking facility to point B, 3.58 miles along the shore from point A, and then along the shore from B to the 
refinery, 
(b) If the per mile cost of underwater construction is p, then C(x) = p\/16 + x 2 + 0.2(9 — x) and 

C(x) = , 0,3 * ^ — 0.2 = gives x c = , ° ,s , which minimizes the construction cost provided x c < 9. The value 

v ; \/l6 + x 2 & c yV-0.04' F c — 

of p that gives x c = 9 miles is 0.218864. Consequently, if the underwater construction costs $218,864 per mile or less, 
then running the pipeline along a straight line directly from the docking facility to the refinery will minimize the cost 
of construction. 

In theory, p would have to be infinite to justify running the pipe directly from the docking facility to point A (i.e., for 
x c to be zero). For all values of p > 0.218864 there is always an x c S (0, 9) that will give a minimum value for C. 
This is proved by looking at C"(x c ) = ^—^ which is always positive for p > 0. 



56. There are two options to consider. The first is to build a new road straight from Village A to Village B. The second is to 
build a new highway segment from Village A to the Old Road, reconstruct a segment of Old Road, and build a new 
highway segment from Old Road to Village B, as shown in the figure. The cost of the first option is Ci = 0.5(150) million 
dollars = 75 million dollars. 



Copyright (c) 2006 Pearson Education 




206 Chapter 4 Applications of Derivatives 



New Construction 




Upgrade 




150-2* 



Old Road 



The construction cost for the second option is C2(x) = 0.5 ( 2\/2500 + x 2 1 + 0.3(150 — 2x) million dollars for 
< x < 75 miles. The following is a graph of C2(x). 




Solving C' 2 (x) 



V2500 - 



0.6 



S 10 15 20 25 30 35 40 45 SO 55 80 65 70 75 
x(milas) 

give x = ± 37.5 miles, but only x = 37.5 miles is in the specified domain. In 



summary, Ci = $75 million, C 2 (0) = $95 million, C 2 (37.5) = $85 million, and C 2 (75) 
a new road straight from village A to village B is the least expensive option. 



).139 million. Consequently, 



57. 



10-Jt D 




The length of pipeline is L(x) = y 4 + x 2 + y 25 + (10 — x) for < x < 10. The following is a graph of L(x) 




Setting the derivative of L(x) equal to zero gives L'(x) 



(10 -x) 



0. Note that 



\/4 + x 2 



cos 9a and 



\A + x 2 ^25 + (10-x) 2 

— — 12^2 = cos 6> B , therefore, L'(x) = when cos A = cos 9 B , or 6 A = 8 B and AACP is similar to ABDP. Use 

\/25 + (10-x) 2 

simple proportions to determine x as follows: I = iSjp =>• x = y w 2.857 miles along the coast from town A to town B. 
If the two towns were on opposite sides of the river, the obvious solution would be to place the pump station on a straight 
line (the shortest distance) between two towns, again forcing #a = #b ■ The shortest length of pipe is the same regardless of 
whether the towns are on thee same or opposite sides of the river. 



Copyright (c) 2006 Pearson Education 




Section 4. 1 Extreme Values of Functions 207 



58. 



50 ft 




30 ft 



(a) The length of guy wire is L(x) = ^900 + x 2 + y/2500 + (150 - x) 2 for < x < 150. The following is a graph of 

L(x). 



Setting L'(x) equal to zero gives L'(x) 




V^OO + x2 ^2500 + (150 -x) 2 



0. Note that 



V900 + X 2 



cos 8a and 



— — - = cos #b- Therefore, L'(x) = when cos 9 a = cos 9b, or 9 a = 9b and A ACE is similar to AABD. 



\/2500 + (150-x) 

Use simple proportions to determine x: ^ 



150 -x 



225 



56.25 feet. 



50 ^ 4 

(b) If the heights of the towers are he and he, and the horizontal distance between them is s, then 



L(x) = Jh 2 c + x 2 + \/h 2 + (s - x) 2 and L'(x) 



(s-x) 



-. However, 



cos 9c and 



i ( s ~ x > = cos #b- Therefore, L'(x) = when cos 9r = cos 9b, or 9q = 9b and A ACE is similar to AABD. 

\/hB + (s-x) 2 



S — X 

he 



*=(^) f 



Simple proportions can again be used to determine the optimum x: £■ 

59. (a) V(x) = 160x - 52x 2 + 4x 3 

V'(x) = 160 - 104x + 12x 2 = 4(x - 2)(3x - 20) 

The only critical point in the interval (0, 5) is at x = 2. The maximum value of V(x) is 144 at x = 2. 
(b) The largest possible volume of the box is 144 cubic units, and it occurs when x = 2 units. 

60. (a) P'(x) = 2 - 200x~ 2 

The only critical point in the interval (0, oo) is at x = 10. The minimum value of P(x) is 40 at x = 10. 
(b) The smallest possible perimeter of the rectangel is 40 units and it occurs at x = 10 units which makes the rectangle a 
10 by 10 square. 



61. Let x represent the length of the base and y 25 — x 2 the height of the triangle. The area of the triangle is represented by 



A(x) = |y25 — x 2 where < x < 5. Consequently, solving A'(x) 



25 - 2x 2 
2^25 -x2 



Since 



A(0) = A(5) = 0, A(x) is maximized at x 



. The largest possible area is A 



( 5 \ _ 25 



-f cm' 



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208 Chapter 4 Applications of Derivatives 



62. (a) From the diagram the perimeter P = 2x 4- 27rr = 400 
=> x = 200 — 7rr. The area A is 2rx 
=> A(r) = 400r - 2?rr 2 where < r < ^. 
(b) A'(r) = 400 - 47rr so the only critical point is r = ™. 
Since A(r) = if r = and x = 200 - irr = 0, the 
values r = — w 31.83 m and x = 100 m maximize the 

7T 

area over the interval < r < — . 



C3 



63. s = -|gt 2 + v t + s => f t = -gt + v = 0=> t = ^.Nows(t) = s <3>t(-§ + v ) = <^> t = Oort 
Thuss(^)=-ig(^) 2 + vo(f)+s 



2vo 
g 

2\'i) 



64. f t = -2sin t + 2cos t, solving | 



=>■ tan t = 1 =4> t 
never negative) =4> the peak current is 2y 2 amps. 



Y + So > so is the maximum height over the interval < t < 

| + nir where n is a nonnegative integer (in this exercise t is 



65. Yes, since f(x) = |x| = yx 2 = (x 2 ) 



2\l/2 



f'(x) 



1 



,2x-l/2 



(2x) 



2 V" I v~v (x2) i/2 | x 
is not required that f be zero at a local extreme point since f ' may be undefined there. 



is not defined at x = 0. Thus it 



66. If f(c) is a local maximum value of f, then f(x) < f(c) for all x in some open interval (a, b) containing c. Since 

f is even, f(— x) = f(x) < f(c) = f(— c) for all — x in the open interval (— b, —a) containing — c. That is, f assumes 
a local maximum at the point — c. This is also clear from the graph of f because the graph of an even function 
is symmetric about the y-axis. 

67. If g(c) is a local minimum value of g, then g(x) > g(c) for all x in some open interval (a, b) containing c. Since 
g is odd, g(— x) = — g(x) < — g(c) = g(— c) for all — x in the open interval (— b, —a) containing — c. That is, g 
assumes a local maximum at the point — c. This is also clear from the graph of g because the graph of an odd 
function is symmetric about the origin. 

68. If there are no boundary points or critical points the function will have no extreme values in its domain. Such 
functions do indeed exist, for example f(x) = xfor— oo<x<oo. (Any other linear function f(x) = mx + b 
with m/0 will do as well.) 

69. (a) f '(x) = 3ax 2 + 2bx + c is a quadratic, so it can have 0, 1, or 2 zeros, which would be the critical points of f. The 



function f(x) 



3x has two critical points atx = —1 and x = 1. The function f(x) = x 3 — 1 has one critical point 



atx = 0. The function f(x) = x 3 + x has no critical points. 





(b) The function can have either two local extreme values or no extreme values. (If there is only one critical point, the 
cubic function has no extreme values.) 



Copyright (c) 2006 Pearson Education 




Section 4. 1 Extreme Values of Functions 209 



70. (a) 




-0.1,0.6] by [-1.5,1.5] 

f(0) = is not a local extreme value because in any open interval containing x = 0, there are infinitely many points 
where f(x) = 1 and where f(x) = — 1. 
(b) One possible answer, on the interval [0, 1]: 
f(x) = j(l-*)cos^, 0<x<l 

This function has no local extreme value at x = 1. Note that it is continuous on [0, 1]. 



7 1 . Maximum value is 11 at x = 5; 

minimum value is 5 on the interval [—3, 2]; 
local maximum at ( — 5, 9) 



72. Maximum value is 4 on the interval [5, 7]; 
minimum value is —4 on the interval [—2, 1]. 



y 

12" 


10 


/ 


\ 8 ' 


/ 


\ 6 


/ 


4 




2 





-6,6] by [0,12] 




[-3,8] by [-5,5] 



73. Maximum value is 5 on the interval [3, oo); 

minimum value is —5 on the interval (— oo, —2]. 




6,6] by [-6,6] 



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210 Chapter 4 Applications of Derivatives 
74. Minimum value is 4 on the interval [— 1, 3] 



\ / 


\ 8 


/ 


\ 6 " 


/ 




/ 


2 









[-6,6] by [0,9] 

75-80. Example CAS commands: 
Maple : 

with(student): 

f := x -> x A 4 - 8*x A 2 + 4*x + 2; 

domain := x=-20/25.. 64/25; 

plot( f(x), domain, color=black, title="Section 4.1 #75(a)" ); 

Df := D(f); 

plot( Df(x), domain, color=black, title="Section 4.1 # 75(b)" ) 

StatPt := fsolve( Df(x)=0, domain ) 

SingPt := NULL; 

EndPt := op(rhs(domain)); 

Pts :=evalf([EndPt,StatPt,SingPt]); 

Values := [seq( f(x), x=Pts )]; 
Maximum value is 2.7608 and occurs at x=2.56 (right endpoint). 
Minimum value J is -6.2680 and occurs at x=l. 86081 (singular point). 
Mathematica : (functions may vary) (see section 2.5 re. RealsOnly ): 

«Miscellaneous "RealOnly" 

Clear[f,x] 

a=-l ; b= 10/3; 

f[x_] =2 + 2x - 3 x 2/3 

f[x] 

Plot[{f[x],f[x]},{x,a,b}] 

NSolve[f [x]==0, x] 

{f[a],f[0],f[x]/.%,f[b]//N 
In more complicated expressions, NSolve may not yield results. In this case, an approximate solution (say 1.1 here) 
is observed from the graph and the following command is used: 

FindRoot[f[x]==0,{x, 1.1}] 

4.2 THE MEAN VALUE THEOREM 



1. When f(x) = x 2 + 2x - 1 for < x < 1, then ^'ig ' = f'(c) =S> 3 = 2c + 2 => c: 

2. When f(x) = x 2 / 3 for < x < 1, then f H±dp. = f'( c ) =^> 1 = (§) c" 1 / 3 => c = ±. 

3. When f(x) = x + 1 for \ < x < 2, then f(2 .^ /2) = f (c) ^0=l-^^>c=l. 

4. When f(x) = yjx - 1 for 1 < x < 3, then ^=fl = f( c ) => & - --'-- - 



2Vc-l 



5. Does not; f(x) is not differentiable at x = in (—1, 8). 



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Section 4.2 The Mean Value Theorem 21 1 



6. Does; f(x) is continuous for every point of [0, 1] and differentiable for every point in (0, 1). 

7. Does; f(x) is continuous for every point of [0, 1] and differentiable for every point in (0, 1). 

8. Does not; f(x) is not continuous at x = because lim_ f(x) = 1 / = f(0). 

x — > 

9. Since f(x) is not continuous on < x < 1, Rolle's Theorem does not apply: lim_ f(x) = lim_ x = 1 

X — > 1 X —> 1 

¥= o = f(i). 



10. Since f(x) must be continuous at x = and x = 1 we have lim f(x) = a = f(0) => a = 3 and 

x^0+ 



lim f(x) = lim f(x) =>• — l+3 + a = m + b =>• 5 = m + b. Since f(x) must also be differentiable at 

x -> 1 x^ 1+ 

x = 1 we have lim f'(x) = lim f'(x) => — 2x + 3| , = ml , 

x-> I" V ' x^ 1+ lx=1 U=1 



1 = m. Therefore, a = 3, m = 1 and b = 4. 



11. (a) i 



n 
iii 

iv 









— • 

-2 


— • 




•— 

2 






*x 


-5 


-4 


-3 


















— •— 


-1 
• 




— • 


2 


— • 


• 


>JC 



(b) Let Xi and r 2 be zeros of the polynomial P(x) = x" + a^x' 1 " 1 + . . . + a x x + a , then P(rj) = P(r 2 ) = 0. 
Since polynomials are everywhere continuous and differentiable, by Rolle's Theorem P'(r) = for some r 
between ri and r 2 , where P'(x) = nx"" 1 + (n — 1) a n .|X"~ 2 + . . . + ai. 

12. With f both differentiable and continuous on [a, b] and f(ri) = f(r 2 ) = ffo) = where ri, r 2 and r3 are in [a, b], 
then by Rolle's Theorem there exists a ci between ri and r 2 such that f'(ci) = and a c 2 between r 2 and r3 
such that f'(c 2 ) = 0. Since f ' is both differentiable and continuous on [a, b], Rolle's Theorem again applies and 
we have a C3 between Ci and c 2 such that f ''(C3) = 0. To generalize, if f has n+ 1 zeros in [a, b] and f w is 
continuous on [a, b], then f W has at least one zero between a and b. 

13. Since f " exists throughout [a, b] the derivative function f ' is continuous there. If f has more than one zero in 
[a, b], say f '(ri) = f'(r 2 ) = for ri ^ r 2 , then by Rolle's Theorem there is a c between ri and r 2 such that 

f "(c) = 0, contrary to f" > throughout [a, b]. Therefore f ' has at most one zero in [a, b]. The same argument 
holds if f" < throughout [a, b]. 

14. If f(x) is a cubic polynomial with four or more zeros, then by Rolle's Theorem f '(x) has three or more zeros, 
f "(x) has 2 or more zeros and f'"(x) has at least one zero. This is a contradiction since f'"(x) is a non-zero 
constant when f(x) is a cubic polynomial. 

15. With f(— 2) = 1 1 > and f(— 1) = — 1 < we conclude from the Intermediate Value Theorem that 
f(x) = x 4 + 3x + 1 has at least one zero between —2 and — 1. Then —2 < x < — 1 => — 8 < x 3 < — 1 

=> -32 < 4x 3 < -4 => -29 < 4x 3 + 3 < -1 => f'(x) < for -2 < x < -1 => f(x) is decreasing on [-2,-1] 
=>• f(x) = has exactly one solution in the interval (—2, —1). 



16. f(x) = x 3 



f'(x) = 3x 2 



> on (—00, 0) =>• f(x) is increasing on (—00, 0). Also, f(x) < if 



x < — 2 and f(x) > if —2 < x < =>• f(x) has exactly one zero in (—00, 0). 



17. g(t)= y/i+ 0+1-4 



f® = * 



2Vt+ 



j— > =>■ g(t) is increasing for t in (0, 00); g(3) = \/3 - 2 < 



and g(15) = v 15 > =>■ g(t) has exactly one zero in (0, 00). 



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212 Chapter 4 Applications of Derivatives 



18. g(t) = t^ + ^iTt - 3.1 => g'(t) = jjijj + ^= > => g(t) is increasing for t in (-1, 1); 
g(-0.99) = -2.5 and g(0.99) = 98.3 => g(t) has exactly one zero in (-1, 1). 

19. r (0) = + sin 2 (§) - 8 =4> r'(0) = 1 + | sin (f) cos (f) = 1 + ± sin (f ) > on (-oo, oo) =4> r(0) is 
increasing on (— oo, oo); r(0) = —8 and r(8) = sin 2 (|) > => r(0) has exactly one zero in (— oo, oo). 

20. r(0) = 20 - cos 2 + \[l =>• r'(0) = 2 + 2 sin cos = 2 + sin 20 > on (-oo, oo) =>■ r(0) is increasing on 
(-oo, oo); r(-27t) = -4?r - cos (-2tt) + \Jl = -4tt - 1 + \fl < and r(2?r) = 4tt-1 + \/2>0 => r(0) has 
exactly one zero in (— oo, oo). 

21. r(0) = sec - i + 5 => r'(0) = (sec 0)(tan 0) + | > on (0, f ) => r(0) is increasing on (0, f ) ; 
r(O.l) ss -994 and r(1.57) « 1260.5 =4> r(0) has exactly one zero in (0, §) . 



22. r(0) = tan - cot - =>• r'(0) = sec 2 + esc 2 - 1 = sec 2 + cot 2 > on (0, f ) 
on (0, § ) ; r (f ) = - \ < and r(1.57) « 1254.2 => r(0) has exactly one zero in (0, §) 



r(0) is increasing 



23. By Corollary 1, f (x) = for all x =>■ f(x) = C, where C is a constant. Since f(— 1) = 3 we have C = 3 

=4> f(x) = 3 for all x. 

24. g(x) = 2x + 5 =>• g'(x) = 2 = f '(x) for all x. By Corollary 2, f(x) = g(x) + C for some constant C. Then 
f(0) = g(0) + C =>• 5 = 5 + C =» C = => f(x) = g(x) = 2x + 5 for all x. 

25. g(x) = x 2 =>• g'(x) = 2x = f'(x) for all x. By Corollary 2, f(x) = g(x) + C. 

(a) f(0) = =5> = g(0) + C = + C => C = =>■ f(x) = x 2 =^ f(2) = 4 

(b) f(l) = => = g(l) + C = 1+ C => C = -1 =>• f(x) = x 2 - 1 => f(2) = 3 

(c) f(-2) = 3 => 3 = g(-2) + C => 3 = 4 + C =^> C = -1 =>■ f(x) = x 2 - 1 =>■ f(2) = 3 

26. g(x) = mx => g'(x) = m, a constant. If f'(x) = m, then by Corollary 2, f(x) = g(x) + b = mx + b where 
b is a constant. Therefore all functions whose derivatives are constant can be graphed as straight lines 

y = mx + b. 



27. (a) y = \ + C 

28. (a) y = x 2 + C 

29. (a) y' = -x~ 2 => y = i + C 



(b) y = f + C 
(b) y = x 2 - x + C 
(b) y = x+i+C 



(c) y = ^ + C 

(c) y = x 3 + x 2 - x + C 

(c) y = 5x-i+C 



30. (a) y' = \ x- 1 / 2 => y = x 1 / 2 +C => y = v ^ + C 
(c) y = 2x 2 -2y / x + C 

31. (a) y = - \ cos2t + C 

(c) y = - 1 cos 2t + 2 sin | + C 



(b) y = 2^+C 



(b) y = 2 sin \ + C 



32. (a) y = tan + C (b) y' = 1 / 2 => y = § 3 / 2 + C 

33. f(x) = x 2 - x + C; = f(0) = 2 - + C => C = => f(x) = x 2 - x 



(c) y = 2 6» 3 / 2 - tan + C 



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Section 4.2 The Mean Value Theorem 213 



34. g(x) = -i+x 2 + C;l=g(-l) 

35. r(0) = 86* + cot 6 + C; = r (f ) : 

=> r(8) = 8(9 + cot (9 - 2tt - 1 



1 



-^ + (-l) 2 + C =* C = -l => g(x) = -l+: 
(|) + cot (f ) + C =>• = 2tt + 1 + C =>• C = -2tt - 1 



36. r(t) = sec t - t + C; = r(0) = sec (0) - + C =4> C = -1 =4> r(t) = sec t - t - 1 



37. v 



38. v 



9.8t + 5 => s = 4.9t 2 + 5t + C; at s = 10 and t = we have C = 10 => s = 4.9t 2 + 5t + 10 
32t - 2 =>• s = 16t 2 - 2t + C; at s = 4 and t = \ we have C = 1 => s = 6t 2 - 2t + 1 



39. v = ^ = sin(7rt) => s = -icos(7rt) + C; at s = and t = we have C = \ =>• s 



1 — COs(7Tt) 



40. v = ^ = |cos(f ) =4> s = sin(|) + C; at s = 1 and t = 7r 2 we have C = 1 => s = sin(f ) + 1 

41. a = 32 => v = 32t + d; at v = 20 and t = we have Ci = 20 => v = 32t + 20 =4> s = 16t 2 + 20t + C 2 ; at s = 5 and 
t = we have C 2 = 5 => s = 16t 2 + 20t + 5 

42. a = 9.8 => v = 9.8t + Ci; at v = -3 and t = we have Ci = -3 => v = 9.8t - 3 => s = 4.9t 2 - 3t + C 2 ; at s = and 
t = we have C 2 = => s = 4.9t 2 - 3t 

43. a = -4sin(2t) =4> v = 2cos(2t) + Ci; at v = 2 and t = we have Ci = => v = 2cos(2t) =>■ s = sin(2t) + C 2 ; at s = -3 
and t = we have C 2 = —3 =^> s = sin(2t) — 3 



44. a = 4cos(-) ^ v = -sin(-) + Ci; at v = and t = we have Ci = 
s = — 1 and t = we have C 2 = => s = — cosf — ) 



\ -sin( -) =^> s = -cos(-) + C 2 ; at 



-sinf-l 

7T V 7T ^ 



45. If T(t) is the temperature of the thermometer at time t, then T(0) = - 19° C and T(14) = 100° C. From the 
Mean Value Theorem there exists a < t < 14 such that T( ^lzl <0) = 8.5° C/sec = T'(t ), the rate at whic 
the temperature was changing at t = to as measured by the rising mercury on the thermometer. 



46. Because the trucker's average speed was 79.5 mph, by the Mean Value Theorem, the trucker must have been going that 
speed at least once during the trip. 

47. Because its average speed was approximately 7.667 knots, and by the Mean Value Theorem, it must have been going that 
speed at least once during the trip. 

48. The runner's average speed for the marathon was approximately 1 1.909 mph. Therefore, by the Mean Value Theorem, the 
runner must have been going that speed at least once during the marathon. Since the initial speed and final speed are both 
mph and the runner's speed is continuous, by the Intermediate Value Theorem, the runner's speed must have been 1 1 mph 
at least twice. 



49. Let d(t) represent the distance the automobile traveled in time t. The average speed over < t < 2 is 
d(2 ^~p (0) . The Mean Value Theorem says that for some < t < 2, d'(t ) 
the speed of the automobile at time t (which is read on the speedometer) 



d( ^:Q (0) . The Mean Value Theorem says that for some < t < 2, d'(t ) = ^I^ - The value d'(t ) is 



50. a(t) = v'(t) = 1.6 => v(t) = 1.6t + C; at (0, 0) we have C = =>■ v(t) = 1.6t. When t = 30, then v(30) = 48 m/sec. 



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214 Chapter 4 Applications of Derivatives 



5 1 . The conclusion of the Mean Value Theorem yields 



b-a 



3 => ^ 2 (^)=a-b =* c 



ab. 



52. The conclusion of the Mean Value Theorem yields \_\ = 2c => c — '' 



2 ' 



53. f'(x) = [cos x sin(x + 2) + sin x cos(x + 2)] — 2 sin(x + 1) cos(x + 1) = sin(x + x + 2) — sin 2(x + 1) 

= sin (2x + 2) — sin (2x + 2) = 0. Therefore, the function has the constant value f(0) = —sin 2 1 w —0.7081 
which explains why the graph is a horizontal line. 

54. (a) f(x) = (x + 2)(x + l)x(x - l)(x - 2) = x 5 - 5x 3 + 4x is one possibility. 

(b) Graphing f(x) = x 5 - 5x 3 + 4x and f'(x) = 5x 4 - 15x 2 + 4 on [-3, 3] by [-7, 7] we see that each x-intercept of 
f'(x) lies between a pair of x-intercepts of f(x), as expected by Rolle's Theorem, 
y 



y = f'(x) 



y = f(x) 




(c) Yes, since sin is continuous and differentiable on ( — oo, oo). 

55. f(x) must be zero at least once between a and b by the Intermediate Value Theorem. Now suppose that f(x) is zero twice 
between a and b. Then by the Mean Value Theorem, f'(x) would have to be zero at least once between the two zeros of 
f(x), but this can't be true since we are given that f'(x) ^ on this interval. Therefore, f(x) is zero once and only once 
between a and b. 



56. Consider the function k(x) = f(x) — g(x). k(x) is continuous 
and differentiable on [a, b], and since k(a) = f(a) — g(a) and 
k(b) = f(b) — g(b), by the Mean Value Theorem, there must 
be a point c in (a, b) where k'(c) = 0. But since 
k'(c) = f'(c) — g'(c), this means that f'(c) = g'(c), and c is a 
point where the graphs of f and g have tangent lines with the 
same slope, so these lines are either parallel or are the same 
line. 




57. Yes. By Corollary 2 we have f(x) = g(x) + c since f'(x) = g'(x). If the graphs start at the same point x = a, 
then f(a) = g(a) =4> c = =>- f(x) = g(x). 



58. Let f(x) = sin x for a < x < b. From the Mean Value Theorem there exists a c between a and b such that 



sin b — sin a 

b-a 



cose =>■ -1< sin £- sina < 1 => 

— b — a — 



I si "b-a" a | ^ l =*" |sinb-sina| < |b-a|. 



59. By the Mean Value Theorem we have ( g _ a = f'(c) for some point c between a and b. Since b — a > and f(b) < f(a), 
we have f(b) - f(a) < => f'(c) < 0. 



60. The condition is that f should be continuous over [a, b]. The Mean Value Theorem then guarantees the 
existence of a point c in (a, b) such that ^ ~ a = f '(c). If f ' is cont 
maximum value on [a, b], and min f < f'(c) < max f ', as required. 



existence of a point c in (a, b) such that ^ _ a = f '(c). If f ' is continuous, then it has a minimum and 



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Section 4.3 Monotonic Functions and the First Derivative Test 215 



61. f'(x) = (1 + x 4 cos x) * => f"(x) = - (1 + x 4 cos x) 2 (4x 3 cos x - x 4 sin x) 



-x 3 (1 + x 4 cos x) (4 cos x — x sin x) < for < x < 0.1 => f'(x) is decreasing when < x < 0.1 

Mi 

o.i 



min f « 0.9999 and max f = 1. Now we have 0.9999 < f(0 -' ) ~ 1 < 1 => 0.09999 < f(0.1) - 1 < 0.1 



=> 1.09999 < f(0.1) < 1.1. 
62. f (x) = (1 - x 4 ) -1 => f"(x) = - (1 - x 4 )~ 2 (-4x 3 ) = -^-3 > for < x < 0.1 => f'(x) is increasing when 

(1 X ) 



f(0.1)-2 

0.1 < f(0.1) - 2 < 0.10001 =*> 2.1 < f(0.1) < 2.10001. 



< x < 0.1 => min f = 1 and max f = 1.0001. Now we have 1 < ' ^ < 1.0001 



63. (a) Suppose x < 1, then by the Mean Value Theorem f(x ^I^ (1) < => f(x) > f(l). Suppose x > 1, then by the 

Mean Value Theorem fJ ^zj 11 > =>■ f(x) > f(l). Therefore f(x) > 1 for all x since f(l) = 1. 
(b) Yes. From part (a), lim f(x) ^ (1) < and lim fiMnIil > q. Since f'(l) exists, these two one-sided 

X — > 1~ X ' X — > 1+ x 

limits are equal and have the value f'(l) =4> f'(l) < and f'(l) > ^ f'(l) = 0. 

64. From the Mean Value Theorem we have ( ^ ~ a (a) = f'(c) where c is between a and b. But f'(c) = 2pc + q = 
has only one solution c = — £- . (Note: p ^ since f is a quadratic function.) 

4.3 MONOTONIC FUNCTIONS AND THE FIRST DERIVATIVE TEST 

1. (a) f'(x) = x(x — 1) =>- critical points at and 1 

(b) f = +++ | | +++ => increasing on (— oo, 0) and (1, oo), decreasing on (0, 1) 

1 

(c) Local maximum at x = and a local minimum at x = 1 

2. (a) f'(x) = (x - l)(x + 2) => critical points at -2 and 1 

(b) f = +++ | | +++ => increasing on (— oo, —2) and (1, oo), decreasing on (—2, 1) 

-2 1 

(c) Local maximum at x = —2 and a local minimum at x = 1 

3. (a) f'(x) = (x - l) 2 (x + 2) =>- critical points at -2 and 1 

(b) f' = | +++ | +++ => increasing on (—2. 1) and (1, oo), decreasing on (—oo, —2) 

-2 1 

(c) No local maximum and a local minimum at x = —2 

4. (a) f'(x) = (x - l) 2 (x + 2) 2 =^> critical points at -2 and 1 

(b) f = +++ | +++ | +++ => increasing on (— oo, —2) U (—2, 1) U (1, oo), never decreasing 

-2 1 

(c) No local extrema 

5. (a) f'(x) = (x - l)(x + 2)(x - 3) => critical points at -2, 1 and 3 

(b) f = | +++ | | +++ => increasing on (—2, 1) and (3, oo), decreasing on (— oo, —2) and (1,3) 

-2 13 

(c) Local maximum at x = 1, local minima at x = —2 and x = 3 

6. (a) f'(x) = (x - 7)(x + l)(x + 5) =^ critical points at -5, -1 and 7 

(b) f = | +++ | I +++ => increasing on (—5, —1) and (7, oo), decreasing on (—oo, —5) and (—1,7) 

-5 -17" 

(c) Local maximum at x = — 1, local minima at x = —5 and x = 7 



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216 Chapter 4 Applications of Derivatives 

7. (a) f'(x) = x~ 1/3 (x + 2) =4> critical points at -2 and 

(b) f = +++ | )( +++ =>• increasing on (— oo, —2) and (0, oo), decreasing on (—2, 0) 

-2 

(c) Local maximum at x = —2, local minimum at x = 

8. (a) f (x) = x~ 1/2 (x - 3) =>■ critical points at and 3 

(b) f = ( | +++ =>■ increasing on (3, oo), decreasing on (0, 3) 

3 

(c) No local maximum and a local minimum at x = 3 



9. (a) g(t) = -t 2 - 3t + 3 => g'(t) = -2t - 3 => a critical point at t 



3 . _/ 



(— oo, — |) , decreasing on (— |, oo) 

(b) local maximum value of g (— |) - 

(c) absolute maximum is =j- at t = — | 

(d) 

g(t) 

g (t) = -t 2 -3t + 3 6- 



f att=-| 




-+ | , increasing on 

-3/2 



10. (a) g(t) = -3t 2 + 9t + 5 =>• g'(t) = -6t + 9 => a critical point at t = | ; g' = ++- 



(— oo, |) , decreasing on (|, oo) 

(b) local maximum value ofg(|) = ^ att 

(c) absolute maximum is ^ at t = | 
(d) 



g(t) = -3t 2 +9t + 5 




3/2 



increasing on 



11. (a) h(x)= -x 3 + 2x 2 

=> h' = 

4/3 

(b) local maximum value of h ( j) 

(c) no absolute extrema 



h'(x) = -3x 2 + 4x = x(4 - 3x) =^ critical points at x = 0, \ 
+++ | , increasing on (0, |) , decreasing on (— oo,0) and (|, 



Yj at x = | ; local minimum value of h(0) = at x = 



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Section 4.3 Monotonic Functions and the First Derivative Test 217 



(d) 



x 3 +2x 2 




12. (a) h(x) = 2x 3 - 18x =>■ h'(x) = 6x 2 - 18 = 6 be + y/3) (x - y/3) => critical points at x = ± \/3 

=>• h' = +++ I I +++, increasing on ( —00, — y 3 ) and ( y3, 00 ) , decreasing on ( — y3, y 3 

-n/3 y^ V y V ' V 

(b) a local maximum is h ( — y 3 I = 12 y 3 at x = — y 3; local minimum is h ( y 3 J = — 12y 3 at x = y 3 

(c) no absolute extrema 

(d) 




2 3 

h(x) = 2x 3 -18x 



13. (a) f(0) = 36 2 - 46 3 => f (0) = 66- 126 2 = 60(1 - 20) => critical points at 6 = 0, \ => f 

increasing on (0, |) , decreasing on (—00, 0) and (|, 00) 

(b) a local maximum is f (|) = | at = |, a local minimum is f(0) = at 6* = 

(c) no absolute extrema 
(d) 



f(0) = 30 2 -40 3 




I+++I - 
1/2 



14. (a) f(0) = 60 - 3 => f'(0) = 6 — 36> 2 = 3 (y^ - 0) (y 7 ^ + 0) =>■ critical points at = ± y^ => 

f ' = I +++ I , increasing on ( — y 2, y 2), decreasing on ( — 00, — y 2) and ( y 2, 00) 

-V2 V2 

(b) a local maximum is f f y 2 J = 4y 2 at 6 = y 2, a local minimum is f f — y 2 j = — 4y 2 at 6 = — y 2 

(c) no absolute extrema 



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218 Chapter 4 Applications of Derivatives 

(d) 



f(e) 



f(e) = 6e-e 3 




♦*- e 



15. (a) f(r) = 3r 3 + 16r =>• f'(r) = 9r 2 + 16 =4> no critical points 
decreasing 

(b) no local extrema 

(c) no absolute extrema 
(d) 

f(r) 
200 ; 

/ f(r) = 3r ! + 16r 



f = +++++, increasing on (— oo, oo), never 




16. (a) h(r) = (r + if => h'(r) = 3(r + 7) 2 => a critical point at r = -7 =$■ h' 

(— oo, —7) U (—7, oo), never decreasing 

(b) no local extrema 

(c) no absolute extrema 

(d) 

h(r) 

10 4 

h(r) = (r + 7) 3 



-++ | +++, increasing on 

-7 




-5 
-10 



17. (a) f(x) = x 4 - 8x 2 + 16 => f (x) = 4x 3 - 16x = 4x(x + 2)(x - 2) =*• critical points at x = and x = ±2 

=^ f ' = | +++ | | +++, increasing on (—2, 0) and (2, oo), decreasing on (— oo, —2) and (0, 2) 

-2 2 

(b) a local maximum is f(0) = 16 at x = 0, local minima are f(±2) = 0atx= ±2 

(c) no absolute maximum; absolute minimum is at x = ±2 

(d) 

f(x) f (x) = x 4 -8x 2 + 16 




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Section 4.3 Monotonic Functions and the First Derivative Test 219 



18. (a) g(x) = x 4 - 4x 3 + 4x 2 => g'(x) = 4x 3 - 12x 2 + 8x = 4x(x - 2)(x - 1) =>■ critical points at x = 0, 1, 2 

=>• g' = | +++ | | +++, increasing on (0, 1) and (2. oo), decreasing on (— oo,0) and (1, 2) 

12 

(b) a local maximum is g(l) = 1 at x = 1, local minima are g(0) = at x = and g(2) = at x = 2 

(c) no absolute maximum; absolute minimum is at x = 0, 2 

(d) 



g(x) = x 4 -4x J +4x 2 




19. (a) H(t) = 1 1 4 — t G => H'(t) = 6t 3 - 6t 5 = 6t 3 (l + t)(l - t) =>■ critical points at t = 0, ±1 

=> H' = +++ | | +++ | , increasing on (— oo, —1) and (0, 1), decreasing on (— 1, 0) and (1, oo) 

-10 1 

(b) the local maxima are H(— 1) = \ at t = — 1 and H(l) = \ at t = 1, the local minimum is H(0) = at t = 

(c) absolute maximum is | at t = ± 1 ; no absolute minimum 
(d) 

H(t) 

1 

1 - 

H(t) = ft 4 -t 6 





20. (a) K(t) = 15t 3 - t 5 =4> K'(t) = 45t 2 - 5t 4 = 5t 2 (3 + t)(3 - t) => critical points at t = 0, ±3 

=> K' = | +++ | +++ | , increasing on (—3,0) U (0, 3), decreasing on (— oo, —3) and (3, oo) 

-3 3 

(b) a local maximum is K(3) = 162 at t = 3, a local minimum is K(— 3) = —162 at t = — 3 

(c) no absolute extrema 

(d) 

K < t > K(t) = 15t'-t 5 
150 ■ 
100 




21. (a) g(x) = xy/i 



x 2 ) 1/2 => g'(x) = (8-x 2 



' 1/2 +*g; 



x 2 )" 1/2 (-2x) 



-2V2 



, -2] and ( 



2,2V2 



2(2 - x)(2 + x) 
'isjl -x) (2^/2 + > 



=> critical points at x = ±2, ± l^fl => g' = ( | +++ | 

-2^ - 2 2 



) , increasing on (—2, 2), decreasing on 

2a/2 



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220 Chapter 4 Applications of Derivatives 



(b) local maxima are g(2) = 4 at x = 2 and g ( — 2 \J 2 ) = at x = — 2 y 2, local minima are g(— 2) = —4 at 



x = -2 and g I 2\/2 J = at x = 2V2 

(c) absolute maximum is 4 at x = 2; absolute minimum is —4 at x = —2 
(d) 



g(x) 



(x) = xV8-x 2 




2V5-X 

, increasing on (0, 4), decreasing on (— oo, 0) 



22. (a) g(x) = xV5-x = x 2 (5 - x) 1 / 2 => g'(x) = 2x(5 - x) 1 / 2 + x 2 (±) (5 - x)- 1/2 (-l) - 5xl "' ' ! 

=> critical points at x = 0, 4 and 5 =>• g' = | + 

4 5 

and (4, 5) 

(b) a local maximum is g(4) = 1 6 at x = 4, a local minimum is at x = and x = 5 

(c) no absolute maximum; absolute minimum is at x = 0, 5 

(d) 

;(x) = x 2 V5-x 




23. (a) f(x) 



x 2 -3 
x-2 



f'(x) 



2x(x-2)-(x 2 -3)(l) _ (x-3)(x-l) 
(x - 2) 2 (x - 2) 2 



f 



-+- 



2 



+- 



critical points at x = 1,3 
increasing on (— oo, 1) and (3, oo), decreasing on (1, 2) and (2, 3), 



12 3 

discontinuous at x = 2 

(b) a local maximum is f(l) = 2 at x = 1, a local minimum is f(3) = 6 at x = 3 

(c) no absolute extrema 

(d) 

f(x) 





B 


. 






6 








4 
2 




f (x) = 

v ; x-2 


/<?2 




\2 


4 6 



24. (a) f(x) = 3^ 

=> f = ++- 







el, x 3x 2 (3x 2 + l)-x 3 (6x) 3x 2 (x 2 + l) ... , ... A 

f (x) = (3x 2 + i) 2 = (3x 2 + i) 2 => a cntlcal P° lnt at x = 

-+, increasing on (— oo, 0) U (0, oo), and never decreasing 



(b) no local extrema 

(c) no absolute extrema 



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Section 4.3 Monotonic Functions and the First Derivative Test 221 



(d) 




-0.5 



25. (a) f(x) = x^ 3 (x + 8) = x 4 / 3 + 8X 1 / 3 => f ( x ) = f x 1 / 3 + f x~ 2 / 3 = ^±^ =>■ critical points at x = 0, -2 

=^ f = | +++)(+++, increasing on (—2, 0) U (0, oo), decreasing on (— oo, —2) 

-2 

(b) no local maximum, a local minimum is f(— 2) = —6 V 2 w —7.56 at x = —2 

(c) no absolute maximum; absolute minimum is —6 y 2 at x = —2 

(d) 



f(x) 
10 + 



-10 



f(x) = x" 3 (x + 8) 



26. (a) g(x) = x 2 / 3 (x + 5) = x 5 / 3 + 5x 2 / 3 => g'(x) = f x 2 / 3 + ^ x" 1 / 3 = 5^ + 2) ^> critical points at x = -2 and 

x = =^ g' = +++ | )(+++, increasing on (— oo, —2) and (0, oo), decreasing on (—2,0) 

-2 

(b) local maximum is g(— 2) = 3 y 4 w 4.762 at x = —2, a local minimum is g(0) = at x = 

(c) no absolute extrema 
(d) 



9(1) 



x + 5 



12 3 




27. (a) h(x) = x 1 / 3 (x 2 - 4) = x 7 / 3 - 4X 1 / 3 => h'(x) = \ x 4 / 3 - f x~ 2 / 3 



/7x + 2 K/7x-2 



3^ 



critical points at 



x -0 i 2 - 



h' 



- + - 



)( 



-+ 



-2/\/7 



2/^7 



increasing on f — oo, —h ) and ( -?-, oo J , decreasing on 



(^,0)and(0,^) 

/ —9 \ 24 \/2 _9 / 9 \ 24 \/2 

(b) local maximum is h I — f- I = T ' 6 « 3.12 at x = —£, the local minimum is h I -4- 1 = 

(c) no absolute extrema 



7V6 



^3.12 



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222 Chapter 4 Applications of Derivatives 

(d) 




h(x) = x"' 3 (x 2 -4) 



28. (a) k(x) = x 2 / 3 (x 2 - 4) = x 8 / 3 - 4x 2 / 3 => k'(x) = § x 5 / 3 - | x" 1 / 3 = 8(x+ 3 1 3 ^" 1) =^> critical points at 

x = 0, ± 1 =>- k' = | +++)( | +++, increasing on (— 1, 0) and (1, oo), decreasing on (— oo. — 1) 

-10 1 

and (0,1) 

(b) local maximum is k(0) = at x = 0, local minima are k ( ± 1) = — 3 at x = ±1 

(c) no absolute maximum; absolute minimum is — 3 at x = ±1 
(d) 




k(x) = x 2 ' 3 (x 2 -4) 



29. (a) f(x) = 2x - x 2 =>• f (x) = 2 - 2x = 2(1 - x) =>• a critical point at x = 1 => f = +- 

f(2) = => a local maximum is 1 at x = 1, a local minimum is at x = 2 
(b) absolute maximum is 1 at x = 1 ; no absolute minimum 

(c) 



f (x) = 2x-x 2 




-] andf(l)= 1, 
2 



30. (a) f(x) = (x + l) 2 => f (x) = 2(x + 1) => a critical point at x = -1 =>• f = | +++ ] and f(-l) = 0, f(0) = 1 

-1 

=>• a local maximum is 1 at x = 0, a local minimum is at x = — 1 
(b) no absolute maximum; absolute minimum is at x = — 1 

(c) 

f(x) 





\ 


5- 






\f(x) = (x + l) 2 


4 - 






\ 


3 - 






\ 


2 - 






\^ 


1, 




-4 


-3 -2 -1 


-1 


1 



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Section 4.3 Monotonic Functions and the First Derivative Test 223 

31. (a) g(x) = x 2 - 4x + 4 =>• g'(x) = 2x - 4 = 2(x - 2) => a critical point at x = 2 =>• g' = [ | +++ and 

12 
g(l) = 1, g(2) = =^ a local maximum is 1 at x = 1, a local minimum is g(2) = at x = 2 
(b) no absolute maximum; absolute minimum is at x = 2 

(c) 

g(x) 



S' 








4 ■ 
3 - 




g(x) = x : -4x + 4 




2 - 








1 ■ 








1 


1 


2 3 


4 



32. (a) g(x) = -x 2 - 6x - 9 =^ g'(x) = -2x - 6 = -2(x + 3) =4> a critical point at x = -3 



= [ +- 

-4 



g(— 4) = — 1, g(— 3) = =>- a local maximum is at x = —3, a local minimum is — 1 at x = —4 

(b) absolute maximum is at x = —3; no absolute minimum 

(c) 

g(x) 




-3 



and 



-10 



33. (a) f(t) = 12t - t 3 => f'(t) = 12 - 3t 2 = 3(2 + t)(2 - t) => critical points at t = ±2 =>• f' = [ | - 

-3 -2 

and f(-3) = -9, f(-2) = - 16, f(2) = 16 => local maxima are -9 at t = -3 and 16 at t = -2, a local 
minimum is — 16 at t = —2 
(b) absolute maximum is 16 at t = 2; no absolute minimum 

(c) 

f(t) 

f(t) = 12t-t 3 



- + 




34. (a) f(t) = t 3 - 3t 2 => f '(t) = 3t 2 - 6t = 3t(t - 2) =>• critical points at t = and t = 2 

=> f = +++ | | +++ ] and f(0) = 0, f(2) = -4, f(3) = =^ a local maximum is at t = and t = 3, a 

2 3 

local minimum is —4 at t = 2 
(b) absolute maximum is at t = 0, 3; no absolute minimum 



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224 Chapter 4 Applications of Derivatives 



(c) 



f(t) 




♦- 1 



35. (a) h(x) 



2x 2 + 4x =>• h'(x) = x 2 - 4x + 4 = (x - 2) 2 =4> a critical point at x = 2 



h' = [ +- 




-++ and 



h(0) = => no local maximum, a local minimum is at x = 
(b) no absolute maximum; absolute minimum is at x = 

(c) 



h(x) 
6 

5 
4 
3 
2 

1 



h(x) = 2x 2 +4x 

v ' 3 



36. (a) k(x) = x 3 + 3x 2 + 3x + 1 =>■ k'(x) = 3x 2 + 6x + 3 = 3(x + l) 2 => a critical point at x = -1 

=>■ k' = +++ | +++ ] and k(— 1) = 0, k(0) = 1 =$■ a local maximum is 1 at x = 0, no local minimum 
-1 

(b) absolute maximum is 1 at x = 0; no absolute minimum 

(c) 

k(x) 
2 



k(x) = x 3 +3x 2 +3x+l 1 



37. (a) f(x)= |-2sin(|; 



f 



2tt/3 



^ f (x) = i - cos (I) , f (x) = => cos (I) = \ =4> a critical point at x = y 

-++ ] and f(0) = 0, f (?f ) = f - v/3, f(27r) = ?r ^> local maxima are at x = and tt 



2tt 



at x = 27r, a local minimum is | — y 3 at x 



2^ 



(b) The graph of f rises when f > 0, falls when f ' < 0, 
and has a local minimum value at the point where f ' 
changes from negative to positive. 




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Section 4.3 Monotonic Functions and the First Derivative Test 225 



„2 . 



38. (a) f(x) = —2 cos x — cos^ x 

X = -7T, 0, 7T => f 



=>■ f (x) = 2 sin x + 2 cos x sin x = 2(sin x)(l + cos x) =>■ critical points at 
| +++ ] and f(— 7r) = 1, f(0) = —3, f(7r) = 1 => a local maximum is 1 at 

— 7T Q 7T 

x = ± 7T, a local minimum is —3 at x = 
(b) The graph of f rises when f ' > 0, falls when f ' < 0, ¥ 

and has local extreme values where f ' = 0. The 
function f has a local minimum value at x = 0, where 
the values of f ' change from negative to positive. 




/(x) = -2cos;r -cos 2 *, -it < x < ir 



39. (a) f(x) = esc 2 x - 2 cot x => f '(x) = 2(csc x)(-csc x)(cot x) - 2 (-esc 2 x) = -2 (esc 2 x) (cot x - 1) 



a critical 



point at x 



=> f = (— I + 

tt/4 n 

(b) The graph of f rises when f ' > 0, falls when f ' < 0. 
and has a local minimum value at the point where 
f ' = and the values of f ' change from negative to 
positive. The graph of f steepens as f'(x) — > ± oo 



) and f (f ) =0 =>• no local maximum, a local minimum is at x 



fix) = csc ! jr -2cotx, < x < n 




40. (a) f(x) = sec 2 x — 2 tan x => f'(x) = 2(sec x)(sec x)(tan x) — 2 sec 2 x = (2 sec 2 x) (tan x — 1) =4> a critical point 



atx=|^f' = ( | +++) andf(f) = 

-7t/2 tt/4 tt/2 

(b) The graph of f rises when f ' > 0, falls when f ' < 0, 

and has a local minimum value where f ' = and the 

values of f ' change from negative to positive. 



no local maximum, a local minimum is at x 



f(x) = sec 2 jr -2 tan* 




41. h(0) = 3 cos (I) => h'(0) = - | sin (§) =^ h' = [ ] , (0,3) and (2tt, -3) =^ a local maximum is 3 at 9 = 0, 



2tt 



a local minimum is — 3 at 6 = 2ir 



42. h(0) = 5 sin (f ) => h'(0) = | cos (f ) =>• h' = [ +++ ] , (0, 0) and (tt, 5) => a local maximum is 5 at 8 = n, a local 



minimum is at 9 = 



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226 Chapter 4 Applications of Derivatives 



43. (a) 



y =/M 



r\ 



lal 



(b) 



y =m 



1 



(c) 




(c) 



(d) 




i 

(d) 



44. (a) 



y - f(x) 



(b) 




(c) 



(d) 





fix! 



45. (a) 



y 






1 


. 


s / 




2 




****/ N**-*- 


y = g(x) 







2 





(b) 




46. (a) 



, 


I 


1- 


f y = h(x) 


-t -J o 


2 4 








47. f(x) = x 3 - 3x + 2 => f (x) = 3x 2 - 3 = 3(x - l)(x + 1) => f = +++ 
f (x) > for x = c = 2. 



+++ =4> rising for x = c = 2 since 



f(x) = ax 2 + bx + c = a (x ; 



(x 2 + ^) + c = a(x 2 + fex+^)-£ + c = a(x+^) 



b \ 2 b 2 - 4ac 



a parabola whose 



4a 2 J 4a ' ""T ' 2ai 4a 

vertex is at x = — ^ . Thus when a > 0, f is increasing on (^, oo) and decreasing on (—00, ^) ; when a < 0, 

f is increasing on (—00. =^) and decreasing on (=£, 00) . Also note that f'(x) = 2ax + b = 2a (x + |-) => for 

a>0, f' = I +++ ; for a < 0, f = +++ | . 

— b/2a -b/2a 



4.4 CONCAVITY AND CURVE SKETCHING 



1. y=^-^-2x+i^>y' = x 2 -x-2 = (x- 2)(x +1) => y" = 2x - 1 = 2 (x - |) . The graph is rising on 
(—00, —1) and (2, 00), falling on (— 1, 2), concave up on (|, 00) and concave down on (—00, |) . Consequently, 
a local maximum is | at x = — 1 , a local minimum is — 3 at x = 2, and ( h , — |) is a point of inflection. 



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Section 4.4 Concavity and Curve Sketching 227 



2. y = si - 2x 2 + 4 => y' = x 3 - 4x = x (x 2 - 4) = x(x + 2)(x - 2) => y" = 3x 2 - 4 = f y^x + 2^ (y^x - 2) . The 
graph is rising on (—2, 0) and (2, oo), falling on (— oo, —2) and (0, 2), concave up on ( — oo, -y- j and I -4= , oo 1 
and concave down on ( — 7" ) ~7~ ) • Consequently, a local maxirr 
x = ± 2, and ( — j- , ^ j and ( -4= , y J are points of inflection. 



-j- j -7- ) • Consequently, a local maximum is 4 at x = 0, local minima are at 



3. y = ! (x 2 - 1) 2/3 => y'=(|)(|)(x 2 -l)- 1/3 (2x) = x(x 2 -l)- 1/3 ,y' = -— ) (+++ | —)(+++ 

-10 1 

=>■ the graph is rising on (—1, 0) and (1, oo), falling on (— oo, —1) and (0, 1) =^ a local maximum is | at x = 0, local 



minima are at x = ± 1; y" = (x 2 - 1) ' + (x) (- i) (x 2 - 1) ' (2x) 



\-^f 



x 2 -3 



3 V(" 2 -!) 4 



+ + + 



-) ( 
-1 



)(• 

1 



- | +++ =4> the graph is concave up on ( — oo, — y 3 J and ( y 3, oo j , concave 
down on I — y 3, y 3 I => points of inflection at ( ± y 3, -j- j 



4. y = 9. X V3 (x 2 _ 7) ^ y'= t L x - 2 / 3 (x 2 -7) + ^x 1 /3(2x)=|x- 2 / 3 (x 2 -l),y' = +++| )( |+++ 

=>■ the graph is rising on (— oo, —1) and (1, oo), falling on (—1, 1) => a local maximum is y at x = — 1, a local 

minimum is - y atx = 1; y" = -x~ 5 / 3 (x 2 - 1) + 3X 1 / 3 = 2x : / 3 + x~ 5 / 3 = x~ 5 / 3 (2x 2 + 1) , 

y" = )( +++ =4> the graph is concave up on (0, oo), concave down on (— oo, 0) => a point of inflection at 


(0,0) 

5. y = x + sin 2x => y' = 1 + 2 cos 2x, y' = [ | +++ | ] => the graph is rising on (— |, |) , falling 

-2tt/3 -tt/3 tt/3 2tt/3 

^ and z _l YJ! 



f,-f)and(f,?f 



local maxima are — y + -y- at x 



at x = | , local minima are 



3 2 at X — 3 """ 3 2 



5 and ^ - ^ at x = y" ; y" = -4 sin 2x, y" = [ 



2tt/3 -tt/2 



- | +++ ] =>■ the 

tt/2 2tt/3 



graph is concave up on (— |, 0) and (|, y) , concave down on (— y, — |) and (0, |) => points of inflection at 
(-f,-f), (0,0), and (|, |) 

6. y = tan x — 4x =S> y' = sec 2 x — 4, y' = ( +++ | | +++ ) => the graph is rising on (— |, — |) and 

-tt/2 -tt/3 tt/3 tt/2 

(|, |) , falling on (— f , f ) => a local maximum is — v 3 + y atx = — | , a local minimum is y 3 - j at x = |; 

y" = 2(sec x)(sec x)(tan x) = 2 (sec 2 x) (tan x), y" = ( | +++ ) =^ the graph is concave up on (0, |) , 

-tt/2 tt/2 

concave down on (— | , 0) => a point of inflection at (0, 0) 



7. If x > 0, sin |x| = sin x and if x < 0, sin |x| = sin(— x) 
= —sin x. From the sketch the graph is rising on 
(- f , - § ) , (0, f ) and (f , 2tt) , falling on (-2tt, - 3 f ) , 
(— |,0) and (|, y) ; local minima are —1 at x = ± y 
and at x = 0; local maxima are 1 at x = ± | and 
at x = ± 2tt; concave up on (— 2ir, — n) and (ir, 2ir), and 
concavedown on {—it, 0) and (0, ir) =$■ points of inflection 
are (— n, 0) and (w, 0) 



(-2R.0) 



(-n/2, 1) 



(-37C/2.-1) 



y = sin|jr|, -2ir sjs2t 
(*/2.D 

(27t,0) 




(3jc/2,-1) 



Cfiojt (c| 1 Pearson Educate Ire, publishing as Pearson Addison-Wesle 



228 Chapter 4 Applications of Derivatives 



y = 2 cos x — v2) 



-2 sin x — v 2, y' = [ 

— 7T 



3n; 
4 ' 



atx 

y" = 



+++ | I +++] =>■ rising on 

■3tt/4 -tt/4 5tt/4 3tt/2 

|) and (^, ^) , falling on (-tt, - 2s) and (- f , ^ ) => local maxima are -2 + tta/2 at x = -tt, y^ + ^ 

4 """ 2 

2cosx, y" = [ +++ I I +++] 

-7r -tt/2 tt/2 3tt/2 



and — :-^ al x 4?, and local minima are - \[7. + ^ at x = _ 3| and _ ^ _ 5^ at x = ^ . 



7T 7T ' 

2' 2; 



concave up on (— it, — |) and (|, y) , concave down on 
points of inflection at('-f,^)and('f,-^f E ) 



9. When y = x 2 - 4x + 3, then y' = 2x - 4 = 2(x - 2) and 
y" = 2. The curve rises on (2, oo) and falls on (— oo, 2). 
At x = 2 there is a minimum. Since y' > 0, the curve is 
concave up for all x. 



>' 






V' 






\ 








J 


_ 






1 2 

1 








-4 -3 -2 -1 




l\ 2 /3 


4 


-1 




(2,-1) 




-2 




Locmin 





10. When y = 6 - 2x - x 2 , then y' = -2 - 2x = -2(1 + x) and 
y" = —2. The curve rises on (-co, —1) and falls on 
(— 1 , oo). At x = — 1 there is a maximum. Since y' < 0, the 
curve is concave down for all x. 




11. When y = x 3 - 3x + 3, then y' = 3x 2 - 3 = 3(x - l)(x + 1) 
and y" = 6x. The curve rises on (— oo, — 1) U (1, oo) and 
falls on (—1, 1). At x = — 1 there is a local maximum and at 
x = 1 a local minimum. The curve is concave down on 
(— oo, 0) and concave up on (0, oo). There is a point of 
inflection at x = 0. 



Loc 


y = x i --ix + i 


max 




( - l - S W 5 


\ 


/ V 


I 


/ Inf 1 1 




/ 2 




/ 1 


' V d.l) 


/ 


Loc min 


1 1 


1 



12. When y = x(6 - 2x) 2 , then y' = -4x(6 - 2x) + (6 - 2x) 2 
= 12(3 - x)(l - x) and y" = -12(3 - x) - 12(1 - x) 
= 24(x — 2). The curve rises on (— oo, 1) U (3, oo) and 
falls on (1, 3). The curve is concave down on (— oo, 2) and 
concave up on (2, oo). At x = 2 there is a point of 
inflection. 




Copyright (c) 2006 Pearson Education 




Section 4.4 Concavity and Curve Sketching 229 



13. When y = -2x 3 + 6x 2 - 3, then y' = -6x 2 + 12x 

= -6x(x - 2) and y" = -12x + 12 = -12(x - 1). The 
curve rises on (0, 2) and falls on (— oo, 0) and (2, oo). 
At x = there is a local minimum and at x = 2 a local 
maximum. The curve is concave up on (— oo, 1) and 
concave down on (1, oo). At x = 1 there is a point of 
inflection. 



(2, 5) Loc max 




y = -2^ + &!''- 3 



14. When y = 1 - 9x - 6x 2 - x 3 , then y' = -9 - 12x - 3x 2 
= -3(x + 3)(x + 1) and y" = -12 - 6x = -6(x + 2). 
The curve rises on (—3, —1) and falls on (— oo, —3) and 
(— 1, oo). At x = — 1 there is a local maximum and at 
x = — 3 a local minimum. The curve is concave up on 
(— oo, —2) and concave down on (—2, oo). At x — —2 
there is a point of inflection. 



y = 1 - 9x - 6jc 2 - x 




15. When y = (x - 2) 3 + 1, then y' = 3(x - 2) 2 and 

y" = 6(x — 2). The curve never falls and there are no 
local extrema. The curve is concave down on (— oo, 2) 
and concave up on (2, oo). At x = 2 there is a point 
of inflection. 





y 




/ 


i 




3 


/ 




2 
1 


Infl / 

(2,1)/ 


-2 - 


1 


/l 2 3 4 




-1 


" / 




-2 


/ ). = ( J :-2) 3 +l 



16. When y = 1 - (x + l) 3 , then y' = -3(x + l) 2 and 
y" = — 6(x + 1). The curve never rises and there are 
no local extrema. The curve is concave up on (— oo, —1) 
and concave down on (—1, oo). At x = 1 there is a 
point of inflection. 




Copyright (c) 2006 Pearson Education 




230 Chapter 4 Applications of Derivatives 



17. When y = x 4 - 2x 2 , then y' = 4x 3 - 4x = 4x(x + l)(x - 1) 

and y" = 12x 2 - 4 = 12 (x + 4?) [x - 4- J . The curve 

rises on (—1,0) and (1, oo) and falls on (— oo, —1) and (0, 1). 
At x = ±1 there are local minima and at x = a local 

maximum. The curve is concave up on ( — oo, \-\ and 



I -T- , oo ) and concave down on 
there are points of inflection. 



v^'vO 



Atx =7I 



t 


y = x 4 -2x 2 


\ - 1 / 

i Loc max f 
\ (0, 0) v / 


-2 \-l J 


V 1 / 2 


Loc min \ /> 

(-1.-D ^7 

(-1/^,-5/9) 

Infl 


^^ yLoc min 
- I^O.-I) 

fl/V3,-5/9) 

Infl 



18. When y = -x 4 + 6x 2 - 4, then y' = -4x 3 + 12x 
= -4x (x + v 7 ^) U - V^) and y" = - 12x2 + 12 
= — 12(x + l)(x — 1). The curve rises on ( —00, — y 3] 
and (0, \/3) , and falls on (-\/3, 0] and (\/3, 00] . At 

x = ± v 3there are local maxima and at x = a local 
minimum. The curve is concave up on (—1, 1) and concave 
down on (—00, — l)and (1, 00). At x = ±1 there are points 
of inflection. 



(- ^".5) 



(/5,5) 




y = -x' + dx 1 - 4 



(0.-4) 



19. When y = 4x 3 - x 4 , then y' = 12x 2 - 4x 3 = 4x 2 (3 - x) and 
y" = 24x — 12x 2 = 12x(2 — x). The curve rises on (—00, 3) 
and falls on (3, 00). At x = 3 there is a local maximum, but 
there is no local minimum. The graph is concave up on 
(0, 2) and concave down on (—00, 0) and (2, 00). There are 
inflection points at x = and x = 2. 



> 


k (3, 27) 




11 




Loc max 




21 


1(2, 16)/ 


I y = 4x 3 


-x 4 


15 


~_ flnfl 






9 








Infl 








(0,0) 3_ 








(\ 


.12 3 


I 4 





20. When y = x 4 + 2x 3 , then y' = 4x 3 + 6x 2 = 2x 2 (2x + 3) and 
y" = 12x 2 + 12x = 12x(x +1). The curve rises on 
(— |, 00) and falls on (—00, — |) . There is a local 
minimum at x = — |, but no local maximum. The curve is 
concave up on (—00, —1) and (0, 00), and concave down on 
(—1,0). At x = — 1 and x = there are points of inflection. 




Cop # (c) 1 Pearson link, k, publishing as Pearson Addison-Wesle 



Section 4.4 Concavity and Curve Sketching 231 



21. When y = x 5 - 5x 4 , then y' = 5x 4 - 20x 3 = 5x 3 (x - 4) and 
y" = 20x 3 — 60x 2 = 20x 2 (x — 3). The curve rises on 
(— oo, 0) and (4, oo), and falls on (0, 4). There is a local 
maximum at x = 0, and a local minimum at x = 4. The 
curve is concave down on (— oo, 3) and concave up on 
(3, oo). At x = 3 there is a point of inflection. 




22. When y = x 
= (|-5 
•5' 



3 (f 
3 (1) 



then y' 
, and y" 



-5 



x(4)(|-5) 3 (i 



'5x 

v 2 



5) 2 (x 



\2 ") \2> 

,■ , y ,. _, ,. M ., ... -J ,.* 4). The curve is rising 
on (— oo, 2) and (10, oo), and falling on (2, 10). There is a 
local maximum at x = 2 and a local minimum at x = 10. 
The curve is concave down on (— oo, 4) and concave up on 
(4, oo). At x = 4 there is a point of inflection. 




23. When y = x + sin x, then y' = 1 + cos x and y" = —sin x. 
The curve rises on (0, 2ir). At x = there is a local and 
absolute minimum and at x = 2tt there is a local and absolute 
maximum. The curve is concave down on (0, tt) and concave 
up on (7r, 27r). At x = 7T there is a point of inflection. 



y 


" Max 

2ir - (2tt,2tt)% 




y = x + sin x / 




(it, tt) y 


■n 


y^iun 




/Min 


I) 


IT 1lT 



24. When y = x — sin x, then y' = 1 — cos x and y" = sin x. 
The curve rises on (0, 2ir). At x = there is a local and 
absolute minimum and at x = 2ir there is a local and absolute 
maximum. The curve is concave up on (0, tt) and concave 
down on (tt, 2ir). At x = 7r there is a point of inflection. 




A v-9/5 
25 A 



25. When y = x 1 / 5 , then y' = | x" 4 / 5 and y" 

The curve rises on (— oo, oo) and there are no extrema. 
The curve is concave up on (— oo, 0) and concave down 
on (0, oo). At x = there is a point of inflection. 




(0,0) 
Infl 



Copyright (c) 2006 Pearson Education 




232 Chapter 4 Applications of Derivatives 



26. When y = x 3 / 5 , then / = | x~ 2 / 5 and y" = - ^ x~ 7 / 5 . 
The curve rises on (— oo, oo) and there are no extrema. 
The curve is concave up on (— oo, 0) and concave down 
on (0, oo). At x = there is a point of inflection. 




27. When y = x 2 / 5 , then y' = § x" 3 / 5 and y" = - ^ x~ 8 / 5 . 
The curve is rising on (0, oo) and falling on (— oo, 0). At 
x = there is a local and absolute minimum. There is 
no local or absolute maximum. The curve is concave 
down on (— oo, 0) and (0, oo). There are no points of 
inflection, but a cusp exists at x = 0. 




28. When y = x 4 / 5 , then y' = f x" 1 / 5 and y" = - ^ x" 6 / 5 . 
The curve is rising on (0, oo) and falling on (— oo, 0). At 
x = there is a local and absolute minimum. There is 
no local or absolute maximum. The curve is concave 
down on (— oo, 0) and (0, oo). There are no points of 
inflection, but a cusp exists at x = 0. 




29. When y = 2x - 3x 2 / 3 , then y' = 2 - 2X" 1 / 3 and 
y" = | x~ 4 / 3 . The curve is rising on (— oo, 0) and 
(1, oo), and falling on (0, 1). There is a local maximum 
at x = and a local minimum at x = 1. The curve is 
concave up on (— oo, 0) and (0, oo). There are no 
points of inflection, but a cusp exists at x = 0. 



Cusp, Loc max 
(0,0) 




30. When y = 5x 2 / 5 - 2x, then y' = 2x~ 3 / 5 -2 = 2 (x^ 3 / 5 - l) 
and y" = — | x -8 / 5 . The curve is rising on (0, 1) and 
falling on (— oo, 0) and (1, oo). There is a local minimum 
at x = and a local maximum at x = 1. The curve is 
concave down on (— oo, 0) and (0, oo). There are no 
points of inflection, but a cusp exists at x = 0. 



(0,0) 



(1,3) 



y = 5x vi - 2x 



Copyright (c) 2006 Pearson Education 




Section 4.4 Concavity and Curve Sketching 233 



31. Wheny = x 2 / 3 (§ 

5 

3 



x) = | x 2/3 - x 5/3 , then 



./ _ 5 Y -l/3 _ 5 Y 2/3 
y — i A i A 



3 



| x - 1 /3(i _ x )and 



nX 



- 4 / 3 (l +2x). 



v" — - 5 y- 4 /3 _ 10 Y -l/3 

y — 9 A 9 A 

The curve is rising on (0, 1) and falling on (— oo, 0) and 
(1, oo). There is a local minimum at x = and a local 
maximum at x = 1. The curve is concave up on (-co, — \) 
and concave down on (— i OJ and (0, oo). There is a point 
of inflection at x = — \ and a cusp at x = 0. 

32. When y = x 2 / 3 (x - 5) = x 5 / 3 - 5x 2 / 3 , then 



y' = § x 2 / 3 



h> x -i/3 = 5 X -1/3( X _ 2 ) a nd 



-1/3 



-4/3 



f x~ 4 / 3 (x + 1). The curve 



y - -g A -r -g- 

is rising on (— oo, 0) and (2, oo), and falling on (0, 2). 
There is a local minimum at x = 2 and a local maximum 
at x = 0. The curve is concave up on (—1, 0) and (0, oo), 
and concave down on (— oo, — 1). There is a point of 
inflection at x = — 1 and a cusp at x = 0. 




(*-5) 




1-1,-6) 



(2.0,-4.76) 



33. When y = x\/$ - x 2 = x (8 - x 2 ) 1/2 , then 



„2\l/2 



)^ + (x)(i)(8-x 2 )- I/ V2x) 



x 2 ) 1/2 (8-2x 2 ) 



2(2 - x)(2 + x) 



and 



2^/2 + xj (2V2-X) 

2\ i to „2\~2C 



-i)(8-x 2 ) 2 (-2x)(8-2x 2 ) + (8-x 2 ) 2 (-4x) 



2x(x 2 -12) 
\/(8-x 2 ) 3 



. The curve is rising on (—2, 2), and falling 



on ( — 2a/ 2, —21 and I 2, 2y 2) . There are local minima 

x = — 2 and x = 2 \J 2, and local maxima at x = — 2y 2 and 
x = 2. The curve is concave up on ( — 2y 2, 0] and 

concave down on 10,2 y 2 J . There is a point of inflection 
at x = 0. 

34. When y = (2 - x 2 ) 3/2 , then y' = (|) (2 - x 2 ) 1/2 (-2x) 

= — 3xy 2 — x 2 = — 3x 
.2*1/2 



2 + x j and 

v" : - ( -3) (2 - x 2 ) I/2 + (-3x) (I) (2 - x 2 )- 1/2 (-2x) 

-6(1-X)(1+X) rj,, ... 

^^-^^^^^— . The curve is rising on 

v/2-x) (%/2 + x) 

— v 2, ) and falling on I 0, v 2 J . There is a local 

maximum at x = 0, and local minima at x = ± \J 2. The 
curve is concave down on (— 1, 1) and concave up on 

I — v 2, — 1 j and ( 1 , \J 2 j . There are points of inflection at 

x= ± 1. 





v 


Loc max 








•»\ <2,4) 




4 


/ 






3 


- / 






2 


- / 




Loc max 








(-2^2, 0) 


1 


7(0,0) 


Infl 


1 2 


1 / 


1 


2 (-2^,0)' A 








Loc min 




1 -3 


- 


y = jcV8-JC 2 




-4 


_ 




(_2,_4)\K 








Loc mill 









(0,2/2") 




(-^",0) 



Copyright (c) 2006 Pearson Education 




234 Chapter 4 Applications of Derivatives 



35. Wheny 



■ 3 



then y' 



2x(x-2)-(x 2 -3)(l) 
x - 2 ' " lk -" 1 3 (x-2) 2 

(x-3)(x-l) A 

(x - 2)2 an<1 

(2x - 4)(x - 2) 2 - (x 2 - 4x + 3)2 (x - 2) _ 



J (x - 2) 4 ~ (x - 2) 3 • 

The curve is rising on (—00, 1) and (3, 00), and falling on 
(1, 2) and (2, 3). There is a local maximum at x = 1 and a 
local minimum at x = 3. The curve is concave down on 
(—00, 2) and concave up on (2, 00). There are no points 
of inflection because x = 2 is not in the domain. 




36. Wheny = 3^, then y 

_ 3x 2 (x 2 + l) 



/ _ 3x 2 (3x 2 +l)-x 3 (6x) 
(3x2 + 1) 2 



(3x2 + 1)2 



and 



„ _ (12x 3 + 6x)(3x 2 + l) -2(3x 2 + l)(6x)(3x 4 + 3x 2 ) 
y ~~ (3x2 + 1) 4 



6x(l-x)(l+x) 



The curve is rising on (—00, 00) so 



(3x2 + l) J 

there are no local extrema. The curve is concave up on 
(—00, —1) and (0, 1), and concave down on (— 1, 0) and 
(1, 00). There are points of inflection at x = — 1, x = 0, 
and x = 1 . 




3x* + l 



37. Wheny = |x 2 - 1| -- 
2x, Ixl > 1 



!. I x l > 1 u 
2 ~ , , then 



<-{ 



and y" 



The 



x 2 , |x| < 1 

2, |x| > 1 
2x, |x| < 1 """ J {-2, |x| < 1 

curve rises on (—1, 0) and (1, 00) and falls on (—00, —1) 
and (0, 1). There is a local maximum at x = and local 
minima at x = ±1. The curve is concave up on (—00, —1) 
and (1, 00), and concave down on (— 1, 1). There are no 
points of inflection because y is not differentiable at x = ±1 
(so there is no tangent line at those points). 




-2 (-1, 0) 

Loc min 



r 



38. Wheny 



2x1 



2x 



- 2x, x < 

x 2 , < x < 2 



then 



^ x 2 - 2x, x > 2 



ti 



x<0 

< x< 2 . 

2, x > 2 



f 2x - 2, x < 
y' = < 2 - 2x, < x < 2 , and y" 

^ 2x - 2, x > 2 ( 

The curve is rising on (0, 1) and (2, 00), and falling on 
(—00, 0) and (1, 2). There is a local maximum at x = 1 and 
local minima at x = and x = 2. The curve is concave up 
on (—00, 0) and (2, 00), and concave down on (0, 2). 
There are no points of inflection because y is not 
differentiable at x = and x = 2 (so there is no tangent 
at those points). 




Copyright (c) 2006 Pearson Education 




Section 4.4 Concavity and Curve Sketching 235 



39. Wheny 




I, 2 V^ 



Since lim y 



-oo and lim y 

x^0+ 

0. There is a local minimum at x 



oo there is a 



cusp at x = 0. There is a local minimum at x = 0, but no 
local maximum. The curve is concave down on (— oo, 0) 
and (0, oo). There are no points of inflection. 



40. Wheny=y^4T= J V^Z> x > 4 

[ a/4-x, x<4 



then 



H5 



, x > 4 
, x < 4 



and y" 



_ (x _ 4) -3/2 



-(4-x)- 



x > 4 
x < 4 



Since lim y' = — oo and lim y' = oo there is a cusp 

x -» 4 x -> 4+ 

at x = 4. There is a local minimum at x = 4, but no local 
maximum. The curve is concave down on (— oo, 4) 
and (4, oo). There are no points of inflection. 



1 1 




2 - ^_^- 

N.V^(ao) 

i ] r .^ i i i i ^ 


-4 -3 


-2 - 


i 


\ 1 2 3 4 
Cusp 
Loc min 




(4,0) 



41. y' = 2 + x - x 2 = (1 + x)(2 - x), y' = | +++ | — 

-1 2 

=>■ rising on (—1, 2), falling on (— oo, —1) and (2, oo) 

=>■ there is a local maximum at x = 2 and a local minimum 

atx= -l;y" = 1 - 2x, y" = +++ | 

1/2 

=£> concave up on (— oo, |) , concave down on Q, oo) 
=>• a point of inflection at x = \ 




42. y' = x 2 - x - 6 = (x - 3)(x + 2), y' = +++ | | ++- 

-2 3 

=> rising on (— oo, —2) and (3, oo), falling on (—2, 3) 
=4> there is a local maximum at x = —2 and a local 

minimum at x = 3; y" = 2x — 1, y" = | +++ 

1/2 

=4> concave up on (|, oo), concave down on (— oo, |) 
=> a point of inflection at x = \ 



x — 2 



43. y' = x(x - 3) 2 , y' = | +++ | +++ =>• rising on 

3 

(0, oo), falling on (—oo, 0) => no local maximum, but there 
is a local minimum at x = 0; y" = (x — 3) 2 + x(2)(x — 3) 

= 3(x - 3)(x - 1), y" = +++ | | +++ => concave 

1 3 

up on (— oo, 1) and (3, oo), concave down on (1, 3) =>• 
points of inflection at x = 1 and x = 3 




Cop # (c) 1 Pearson Etation, k, publishing as Pearson Addison-Wesle 



236 Chapter 4 Applications of Derivatives 

44. y' = x 2 (2 - x), y' = +++ | +++ | =4> rising on 

2 

(— oo, 2), falling on (2, oo) => there is a local maximum at 
x = 2, but no local minimum; y" = 2x(2 — x) + x 2 (— 1) 

= x(4 — 3x), y" = | +++ | => concave up 

4/3 



on (0, |) , concave down on (— oo, 0) and (|, oo) 
of inflection at x = and x = i 



points 




45. y' = x (x 2 - 12) = x (x - 2^) (x + 2^3) , 

y' = - 



-2JZ 



-+|— I +- 
2^3 



rising on 



2 a/3, o) and (2^, 00) , falling on (-00, -2^/t\ 
and I 0, 2 v 3 ) =>■ a local maximum at x = 0, local minima 

at x = ± 2v/3 ; y" = (1) (x 2 - 12) + (x)(2x) 

= 3(x - 2)(x + 2), y" = +++ I I +++ 

-2 2 

=$■ concave up on (—00, —2) and (2, 00), concave down on 

(—2,2) =^ points of inflection at x = ±2 




46. y' = (x - l) 2 (2x + 3), y' = | +++ | +++ 

-3/2 1 

=>- rising on (— |, 00) , falling on (—00, — |) =4> no local 

maximum, a local minimum at x = — | ; 



y" = 2(x - l)(2x + 3) + (x - 1) 2 (2) = 2(x - l)(3x + 2), 



+- 



-2/3 1 



-+ 



concave up on 



^ — 00, — |) and (1, 00), concave down on (— |, l) 
=> points of inflection at x = — I and x = 1 



x-1 



'X--2/3 



X--3/2 



47. y' = (8x - 5x 2 ) (4 - x) 2 = x(8 - 5x)(4 - x) 2 , 

y' = | +++ | | =4> rising on (0, 



a local maximum at 



-I+++I — I- 
8/5 4 

falling on (—00, 0) and (I, 00) 

x = I , a local minimum at x = 0; 

y" = (8 - 10x)(4 - x) 2 + (8x - 5x 2 ) (2)(4 - x)(-l) 

= 4(4 - x) (5x 2 - 16x + 8) , 

y" = +++ | +++ | =► concave up 

8-2^/6 8+2^/6 4 

5 5 

on I —00, ~g J and I + ^ , 4 J , concave down on 
-\* x 2 ^ and (4 j00 ) ^ points of inflection at 
and x = 4 



5 ' 5 

8±2\/6 




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Section 4.4 Concavity and Curve Sketching 237 



y' = (x 2 - 2x) (x - 5) 2 = x(x - 2)(x - 5) 2 , 

y' = +++ | | +++ | +++ => rising on (-00, 0) 

2 5 

and (2, 00), falling on (0, 2) =4> a local maximum at x = 0, 
a local minimum at x = 2; 

y" = (2x - 2)(x - 5) 2 + (x 2 - 2x) (2)(x - 5) 
= 2(x-5)(2x 2 -8x + 5), 



-+- 



4-V6 



4-j-yjj 

2 



-+ =>- concave up 



( — ^—, + 2 J and (5, 00), concave down on 



4-^6 



) and I + ,/ , 5 ) => points of inflection at 



x = i±y^ and x = 5 



x »0 



x«2 



x = 5 




7T 7T \ 
2> 2/ 



y = sec J x, y = ( +++ ) =>- rising on 

-tt/2 tt/2 

never falling =£• no local extrema; 
y" = 2(sec x)(sec x)(tan x) = 2 (sec 2 x) (tan x), 

y" = ( I +++ ) => concave up on (0, | 

-tt/2 tt/2 

concave down on (— § , 0) , is a opoint of inflection 




50. y' = tan x, y' = ( | +++ ) =4> rising on (0, §) , 

-tt/2 tt/2 

falling on (— |, 0) =>- no local maximum, a local minimum 

at x = 0; y" = sec 2 x, y" = ( +++ ) =4> concave up 

-tt/2 tt/2 

on (— |, I) => no points of inflection 




51. y' = cot f , y' = ( +++ I ) => rising on (0,tt), 

T 2tt 

falling on (7r, 27r) =4- a local maximum at = 7r, no local 

minimum; y" = — \ esc 2 f , y" = ( ) => never 

2tt 

concave up, concave down on (0, 2ir) => no points of 
inflection 




52. y' = esc 2 I , y 1 = ( +++ ) =4> rising on (0, 2*7r), never 
2tt 

falling =S> no local extrema; 

y" = 2(csc" 



v -csc 5 J (cot 



~ 2 f] 



2; 



f 2tt 

concave up on (7r, 27t), concave down on (0, 7r) 
a point of inflection at 9 = n 




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238 Chapter 4 Applications of Derivatives 



53. y' = tan 2 6-1 = (tan 9 - l)(tan 9 + 1), 

y' = ( +++| | +++) =4> rising on 

-tt/2 -7r/4 tt/4 tt/2 

(-§,-£) and (f,§), falling on (-£,f) 



=^> a local maximum at 
y" = 2 tan 6> sec 2 9, y" = ( 



a local minimum at 9 



— I+++) 
-tt/2 tt/2 

concave up on (0, |) , concave down on (— |, Oj 

a point of inflection at 9 = 




54. y' = 1 - cot 2 6> = (1 - cot 6>)(1 + cot 0), 

y' = ( I +++I ) => rising on (|, 3 f) 

tt/4 3tt/4 w 

falling on (0, |) and (x> 7r ) =^ a l° ca l maximum at 

6> = ^, a local minimum at 9 = | ; 

y" = -2(cot 61) (-esc 2 9), y" = ( +++ | ) 

tt/2 7r 

=^> concave up on (0, |) , concave down on (|, 7r) 

=4> a point of inflection at # = | 



e=3it/4 




rising on 



55. y' = cost, y' = [+++ | | +++] 

tt/2 3tt/2 2tt 

(0, §) and (^f , 27r) , falling on (§, ^) => local maxima at 

t = | and t = 27T, local minima at t = and t = 4r ; 



y" = -sint, y" = [ | - 

1" 



2tt 



=> concave up on (tt, 2ir), concave down 
on (0, 7r) => a point of inflection at t = 7r 



56. y' = sin t, y' = [ +++ | ] =4> rising on (0, it), 

tt 2tt 

falling on (tt, 2ir) =$■ a local maximum at t = n, local 
minima at t = and t = 27r; y" = cos t, 

y" = [+++| | +++] ^ concave up on (0, |) 

tt/2 3tt/2 2tt 



and 



, 27T] , concave down on 



2' 2 



points 



of inflection at t = | and t = y 




t-3it/2 




57. y' = (x+l) 



-2/3 



- + +) (+- 
-1 



rising on 



(— oo, oo), never falling => no local extrema; 

y" = - | ( X + l)-5/3 ; y " = +++ ) ( 

-1 
=>- concave up on (— oo, —1), concave down on (—1, oo) 
=^ a point of inflection and vertical tangent at x = — 1 



Infl 
Veit tan 



Cop # (c) 1 Pearson Education, k, publishing as Pearson Addison-Wesle 



Section 4.4 Concavity and Curve Sketching 239 



58. y' = (x - 2)- 1 / 3 , y' = )(• 



rising on (2, oo), 



falling on (— oo, 2) => no local maximum, but a local 

minimum at x = 2; y" = — | (x — 2)~ 4 / 3 , 

y" = )( => concave down on (— oo,2) and 

2 
(2, oo) => no points of inflection, but there is a cusp at 

x = 2 




59. y' = x~ 2 / 3 (x - 1), y' = )( | +++ =>• rising on 

1 

(1, oo), falling on (—oo, 1) =4> no local maximum, but a 

local minimum at x = 1; y" = = x -2 / 3 + \ x~ 5 / 3 

= \ x- 5 / 3 (x + 2), y" = +++ | )( +++ 

-2 

=£> concave up on (—oo, —2) and (0, oo), concave down on 
(—2,0) =^ points of inflection at x = —2 and x = 0, and a 
vertical tangent at x = 




60. y' = x" 4 / 5 (x + 1), y' = | +++)(+++ =>• rising on 

-1 

(—1,0) and (0, oo), falling on (—oo, —1) => no local 
maximum, but a local minimum at x = — 1 ; 

y" = I x-1/5 _ 4 x -9/5 = 1 x -9/5 (x _ 4)> 

y" = +++ )( | +++ => concave up on (—00, 0) and 

4 

(4, 00), concave down on (0, 4) =>■ points of inflection at 
x = and x = 4, and a vertical tangent at x = 




61. y' 



-2x, x < , 
2x, x > ' y 







rising on 



(—00, 00) => no local extrema 



-"={t 



x<0 
x>0 ' 

y" = )( +++ => concave up on (0, 00), concave 


down on (—00,0) => a point of inflection at x = 




62. y' 



-x 2 , x < , 
x 2 , x > ' y 







rising on 



(0, 00), falling on (—00, 0) => no local maximum, but a 

, • • ^ 11 \ ~2x, x < 

local minimum at x = 0; y = < „ „, 

J \ 2x, x > 



+- 







concave up on (—00, 00) 



no point of inflection 




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240 Chapter 4 Applications of Derivatives 



63. The graph of y = f "(x) =>- the graph of y 
up on (0, oo), concave down on (—oo,0) - 
inflection at x = 0; the graph of y = f '(x) 

=>■ y 1 = +++ | | +++ => the graph y 

both a local maximum and a local minimum 



f(x) is concave 
a point of 



f(x) has 




64. The graph of y = f"(x) =>• y" = +++ | =* the 

graph of y = f(x) has a point of inflection, the graph of 

y = f'( x ) => y' = | +++ | =4> the graph of 

y = f(x) has both a local maximum and a local minimum 




65. The graph of y = f"(x) =► y" = | +++ | 

=4> the graph of y = f(x) has two points of inflection, the 

graph of y = f'(x) => y' = | +++ =>■ the graph of 

y = f(x) has a local minimum 



y 






> . 






p 






^— y 




N^Infl 


In il*C 


nX 














Loc N. 








rain 


y" 



66. The graph of y = f"(x) ^ y" = +++ | ^> the 

graph of y = f(x) has a point of inflection; the graph of 

y = f'( x ) =>■ y' = I +++ I => the graph of 

y = f(x) has both a local maximum and a local minimum 




67. 



Point 


y' 


y" 


P 


- 


+ 


Q 


+ 





R 


+ 


- 


S 





- 


T 


- 


- 



Copyright (c) 2006 Pearson Education 




Section 4.4 Concavity and Curve Sketching 241 



68. 



69. 














7 - 




(6,7) 




\ 4 

1 


JXV 




(4,4) 







2 


4 


6 





70. 




71. Graphs printed in color can shift during a press run, so your values may differ somewhat from those given here. 

(a) The body is moving away from the origin when |displacement| is increasing as t increases, < t < 2 and 

6 < t < 9.5; the body is moving toward the origin when |displacementj is decreasing as t increases, 2 < t < 6 
and 9.5 < t < 15 

(b) The velocity will be zero when the slope of the tangent line for y = s(t) is horizontal. The velocity is zero 
when t is approximately 2, 6, or 9.5 sec. 

(c) The acceleration will be zero at those values of t where the curve y = s(t) has points of inflection. The 
acceleration is zero when t is approximately 4, 7.5, or 12.5 sec. 

(d) The acceleration is positive when the concavity is up, 4 < t < 7.5 and 12.5 < t < 15; the acceleration is 
negative when the concavity is down, < t < 4 and 7.5 < t < 12.5 

72. (a) The body is moving away from the origin when |displacement| is increasing as t increases, 1.5 < t < 4, 

10 < t < 12 and 13.5 < t < 16; the body is moving toward the origin when | displacement! i s decreasing as t 
increases, < t < 1.5, 4 < t < 10 and 12 < t < 13.5 

(b) The velocity will be zero when the slope of the tangent line for y = s(t) is horizontal. The velocity is zero 
when t is approximately 0, 4, 12 or 16 sec. 

(c) The acceleration will be zero at those values of t where the curve y = s(t) has points of inflection. The 
acceleration is zero when t is approximately 1.5, 6, 8, 10.5, or 13.5 sec. 

(d) The acceleration is positive when the concavity is up, < t < 1.5, 6 < t < 8 and 10 < t < 13.5, the 
acceleration is negative when the concavity is down, 1.5 < t < 6, 8 < t < 10 and 13.5 < t < 16. 



73. The marginal cost is ^ which changes from decreasing to increasing when its derivative *p is zero. This is a 
point of inflection of the cost curve and occurs when the production level x is approximately 60 thousand units. 



dy 



74. The marginal revenue is -^ and it is increasing when its derivative -A is positive => the curve is concave up 



d 2 y 



< t < 2 and 5 < t < 9; marginal revenue is decreasing when jp < 
2 < t < 5 and 9 < t < 12. 



the curve is concave down 



75. When y' = (x - l) 2 (x - 2), then y" = 2(x - l)(x - 2) + (x - l) 2 . The curve falls on (-oo, 2) and rises on 

(2, oo). At x = 2 there is a local minimum. There is no local maximum. The curve is concave upward on (— oo, 1) and 

Copyright (c) 1 ton E Won, Inc., puiing as Pearson Addison-Wesle 



242 Chapter 4 Applications of Derivatives 

(|, oo) , and concave downward on (l, |) . At x = 1 or x = | there are inflection points. 

76. When y' = (x - l) 2 (x - 2)(x - 4), then y" = 2(x - l)(x - 2)(x - 4) + (x - l) 2 (x - 4) + (x - l) 2 (x - 2) 

= (x - 1) [2 (x 2 - 6x + 8) + (x 2 - 5x + 4) + (x 2 - 3x + 2)] = 2(x - 1) (2x 2 - lOx + 11). The curve rises on 
(— oo, 2) and (4, oo) and falls on (2, 4). At x = 2 there is a local maximum and at x = 4 a local minimum. The 

curve is concave downward on (— oo, 1) and I ~£ , ~y 1 and concave upward on ( 1, ~£ I and 



I ^ , oo ) . At x = 1, — ^— and ~y there are inflection points. 



77. The graph must be concave down for x > because 
f"(x) = - 4 < 0. 




78. The second derivative, being continuous and never zero, cannot change sign. Therefore the graph will always 
be concave up or concave down so it will have no inflection points and no cusps or corners. 



79. The curve will have a point of inflection at x = 1 if 1 is a solution of y" = 0; y = x 3 + bx 2 + ex + d 



3x 2 + 2bx + c 



6x + 2band6(l) + 2b = 



80. (a) True. If f(x) is a polynomial of even degree then f ' is of odd degree. Every polynomial of odd degree has 
at least one real root =>■ f '(x) = for some x = r => f has a horizontal tangent at x = r. 
(b) False. For example, f(x) = x — 1 is a polynomial of odd degree but f '(x) = 1 is never 0. As another 

example, y = I x 3 + x 2 + x is a polynomial of odd degree, but y' = x 2 + 2x + 1 = (x + l) 2 > for all x. 



81. (a) f(x) = ax 2 + bx + c = a (x 2 + \ x) + c = a (x 2 + \ x + ^) - g + c = a (x + ^f 
whose vertex is at x = — £■ =>- the coordinates of the vertex are ( — j- , — b 4a 4ac ) 



b 2 -4ac 
4a 



a parabola 



(b) The second derivative, f "(x) = 2a, describes concavity 
when a < the parabola is concave down. 



when a > the parabola is concave up and 



82. No, f"(x) could be decreasing to zero at x = c and then increase again so it would be concave up on every 
interval even though f"(x) = 0. For example f(x) = x 4 is always concave up even though f"(0) = 0. 

83. A quadratic curve never has an inflection point. If y = ax 2 + bx + c where a^0, then y' = 2ax + b and 
y" = 2a. Since 2a is a constant, it is not possible for y" to change signs. 



84. A cubic curve always has exactly one inflection point. If y = ax 3 + bx 2 + ex + d where a ^ 0, then 

y' = 3ax 2 + 2bx + c and y" = 6ax + 2b. Since ^ is a solution of y" = 0, we have that y" changes its sign 



at x = — j- and y' exists everywhere (so there is a tangent at x = — £-). Thus the curve has an inflection 
point at x = — ^- . There are no other inflection points because y" changes sign only at this zero. 



Copyright (c) 2006 Pearson Education 




Section 4.4 Concavity and Curve Sketching 243 



85. If y = x 5 - 5x 4 - 240, then y' = 5x 3 (x - 4) and 
y" = 20x 2 (x — 3). The zeros of y' are extrema, and 
there is a point of inflection at x = 3. 



v 
200 - 


y 

i 


= 20x 2 (j:-3)/ / 




-200 






^ -X^3 A 

/=5.t 3 (*-4) 
5x 4 - 240 s v_y 


5 


-400 


y- 


=* 5 - 





86. If y = x 3 - 12x 2 , then y' = 3x(x - 8) and 
y" = 6(x — 4). The zeros of y' and y" are 
extrema and points of inflection, respectively. 



: 

N. 50 


y' 


-6(x-41^ 


"iU^J 


^2*^ 4 


i^S 


/^50 
-100 




y' » 3x(x - 8) 


-150 
-200 




\y-x 3 -12x 2 


-250 







87. If y = f x 5 + 16x 2 - 25, then y' = 4x (x 3 + 8) and 
y" = 16 (x 3 + 2) . The zeros of y' and y" are 
extrema and points of inflection, respectively. 



, y 


\ 100 




\ 50 




-3/ \y_>^ 


_^S' 2 3 '" 


/ / ~ 50 


- > = ijc 5 + 16x 2 -25 


' / -100 




"=16(x 3 + 2) 





88. Ify 



r - f -4x 2 + 12x + 20, then 
y' = x 3 - x 2 - 8x + 12 = (x + 3)(x - 2) 2 . 
So y has a local minimum at x = — 3 as its only extreme 
value. Also y" = 3x 2 - 2x - 8 = (3x + 4)(x - 2) and there 
are inflection points at both zeros, — | and 2, of y". 



y"-(3x + 4)(x-2) 



y-T-, -4X..+ 12X + 20 



21. 




The graph of f falls where f ' < 0, rises where f ' > 0, 
and has horizontal tangents where f = 0. It has local 
minima at points where f ' changes from negative to 
positive and local maxima where f ' changes from 
positive to negative. The graph of f is concave down 
where f " < and concave up where f " > 0. It has an 
inflection point each time f "changes sign, provided a tangent 
line exists there. 



y 
f\ r\ rl / /f 

f / 



Copyright (c) 2006 Pearson Education 




244 Chapter 4 Applications of Derivatives 



90. The graph f is concave down where f " < 0, and concave 
up where f " > 0. It has an inflection point each time 
f " changes sign, provided a tangent line exists there. 




91. (a) It appears to control the number and magnitude of the 

local extrema. If k < 0, there is a local maximum to the 
left of the origin and a local minimum to the right. The 
larger the magnitude of k (k < 0), the greater the 
magnitude of the extrema. If k > 0, the graph has only 
positive slopes and lies entirely in the first and third 
quadrants with no local extrema. The graph becomes 
increasingly steep and straight as k — ► oo. 




(b) f '(x) = 3x 2 + k =>■ the discriminant 2 
negative for k > 0; f ' has two zeros x = 

when k > 0; the sign of k controls the number of local extrema 

(c) As k — > oo, f'(x) — > oo and the graph becomes 
increasingly steep and straight. As k — > — oo, the 
crest of the graph (local maximum) in the second 
quadrant becomes increasingly high and the trough 
(local minimum) in the fourth quadrant becomes 
increasingly deep. 



- 4(3)(k) = — 12k is positive for k < 0, zero for k = 0, and 

± J — | when k < 0, one zero x = when k = and no real zeros 



y 




, 200 




N. \100 


J/ 


~y v^ 5 " ° 


^?\— 2~~" k/ 


/ / 

/ -100 


\v_ 


-200 





92. (a) It appears to control the concavity and the number of 
local extrema. 




k=-10 



Copyright (c) 2006 Pearson Education 




Section 4.5 Applied Optimization Problems 245 



(b) f(x) = x 4 + kx 3 + 6x 2 => f (x) = 4x 3 + 3kx 2 + 12x 

=> f"(x) = 12x 2 + 6kx +12 => the discriminant is 

36k 2 - 4(12)(12) = 36(k + 4)(k - 4), so the sign line 

of the discriminant is +++ | | +++ =>• the 

-4 4 

discriminant is positive when |k| > 4, zero when 

k = ± 4, and negative when |k| < 4; f"(x) = has 

two zeros when |k| > 4, one zero when k = ± 4, and 

no real zeros for |k| < 4; the value of k controls the 

number of possible points of inflection. 




93. (a) If y = x 2 / 3 (x 2 - 2) , then y' = f x" 1 / 3 (2x 2 - 1) and 
y" = | x~ 4 ' 3 (10x 2 + 1) . The curve rises on 

(- 4j , OJ and ( X , ooj and falls on (-co, - -^j 



and 



The curve is concave up on (— oo, 0) 



and (0, oo ). 



(b) A cusp since lim y' = oo and lim y' 

x -» x -» 0+ 




y=X 1 '\x 1 -2) 



94. (a) If y = 9x 2 / 3 (x - 1), then y' 

„ _ 10 (x+l) 



15 (x- 



xV3 



and 



X 1 3 



The curve rises on (— oo, 0) and 
I, oo) and falls on (0, 1) . The curve is concave 



down on 
(0,oo). 



-oo, - 



i) and concave up on (— 1, 0) and 



(b) A cusp since lim y' 

x — > 



oo and lim y' 

x^0+ 




95. Yes: y = x + 3 sin 2x =4> y' = 2x + 6 cos 2x. The graph 
of y' is zero near —3 and this indicates a horizontal tangent 
near x = —3. 




2x + 6 cos 2x 



4.5 APPLIED OPTIMIZATION PROBLEMS 



32 _ 2 (I 2 -16) 



1. Let £ and w represent the length and width of the rectangle, respectively. With an area of 16 in. 2 , we have 
that (f)(w) = 16 => w = let 1 => the perimeter is P = U + 2w = 2£ + 12tr l and P'(£) = 2 
Solving P'(£) = =^ 2(f + 4K£-4) = Q ^ £ = _4 5 4 since £ > o for the length of a rectangle, I must be 4 and 
w = 4 =$■ the perimeter is 16 in., a minimum since P"(£) = || > 0. 



2. Let x represent the length of the rectangle in meters (0 < x < 4) Then the width is 4 — x and the area is 

A(x) = x(4 - x) = 4x - x 2 . Since A'(x) = 4 - 2x, the critical point occurs at x = 2. Since, A'(x) > for < x < 2 and 
A'(x) < for 2 < x < 4, this critical point corresponds to the maximum area. The rectangle with the largest area measures 
2 m by 4 — 2 = 2 m, so it is a square. 



Copyright (c) 2006 Pearson Education 




246 Chapter 4 Applications of Derivatives 
Graphical Support: 



A(x) 




(a) The line containing point P also contains the points (0, 1) and (1, 0) => the line containing P is y = 1 — x 

=> a general point on that line is (x, 1 — x). 

(b) The area A(x) = 2x(l — x), where < x < 1. 

(c) When A(x) = 2x - 2x 2 , then A'(x) = => 2-4x = => x = ± . Since A(0) = and A(l) = 0, we conclude 
that A (i) = | sq units is the largest area. The dimensions are 1 unit by | unit. 



The area of the rectangle is A = 2xy = 2x ( 1 2 — x 2 ) , 
where < x < ^f\2 . Solving A'(x) = =4> 24 - 6x 2 = 
=4> x = —2 or 2. Now —2 is not in the domain, and since 

A(0) = and A (\f\2\ = 0, we conclude that A(2) = 32 



square units is the maximum area. The dimensions are 4 units 
by 8 units. 



/\ 



a 



\ 



The volume of the box is V(x) = x(15 — 2x)(8 — 2x) 
= 120x - 46x 2 + 4x 3 , where < x < 4. Solving V'(x) 
=>■ 120 - 92x + 12x 2 = 4(6 - x)(5 - 3x) = => x = 
or 6, but 6 is not in the domain. Since V(0) = V(4) = 0, 
V (|) = 2 |ip ~ 91 in 3 must be the maximum volume of 
the box with dimensions yX fx ? inches. 



15-2X 









1 

1 













6. The area of the triangle is A = | ba = | y 400 — b 2 , where 
< b < 20. Then ^ = ± V '400 - b 2 , h " 

- db 2 v 2\/40()-b2 

= 2 P°' h \ = =^> the interior critical point is b = 10\/2. 

V400 - b 2 f V 



When b = or 20, the area is zero => A I 10 v 2 J is the 
maximum area. When a 2 + b 2 = 400 and b = 10y 2, the 



value of a is also 10 y 2 
a = b. 



the maximum area occurs when 



(0.b) . 

\20 

(«.0) 



7. The area is A(x) = x(800 - 2x), where < x < 400. 
Solving A'(x) = 800 - 4x = => x = 200. With 
A(0) = A(400) = 0, the maximum area is 
A(200) = 80,000 m 2 . The dimensions are 200 m by 400 m. 



rlvsr 



800-2X 



Copyright (c) 2006 Pearson Education 




Section 4.5 Applied Optimization Problems 247 



The area is 2xy = 216 
needed is P = 4x + 3y 
g=4-ff=0=>x 



=> y = 

= 4x4 

2 S j 



— . The amount of fence 

X 

324x _1 , where < x; 
= => the critical points are 



and ± 9, but and —9 are not in the domain. Then 
P"(9) > =4> at x = 9 there is a minimum =>- the 
dimensions of the outer rectangle are 18 m by 12 m 
=> 72 meters of fence will be needed. 



(a) We minimize the weight = tS where S is the surface area, and t is the thickness of the steel walls of the tank. The 

its di 
?200). Treating the 



surace area is S = x 2 + 4xy where x is the length of a side of the square base of the tank, and y is its depth. The 



volume of the tank must be 500ft 3 => y = ™ . Therefore, the weight of the tank is w(x) = t(x 2 
thickness as a constant gives w'(x) = t(2x — ^2) for x.0. The critical value is at x = 10. Since 
w"(10) = t(2 + ^gr) > 0, there is a minimum at x = 10. Therefore, the optimum dimensions of the tank are 10 ft on 
the base edges and 5 ft deep, 
(b) Minimizing the surface area of the tank minimizes its weight for a given wall thickness. The thickness of the steel 
walls would likely be determined by other considerations such as structural requirements. 



10. (a) The volume of the tank being 1125 ft 3 , we have that yx 2 = 1125 



c(x) = 5x 2 + 30x(±±P), where < x. Then c'(x) = lOx 



X 2 



ii|2 . The cost of building the tank is 
■ the critical points are and 15, but is not 



in the domain. Thus, c"(15) > => at x = 15 we have a minimum. The values of x = 15 ft and y = 5 f t will 
minimize the cost, 
(b) The cost function c = 5(x 2 + 4xy) + lOxy, can be separated into two items: (1) the cost of the materials and labor to 
fabricate the tank, and (2) the cost for the excavation. Since the area of the sides and bottom of the tanks is (x 2 + 4xy), 
it can be deduced that the unit cost to fabricate the tanks is $5/ft 2 . Normally, excavation costs are per unit volume of 
excavated material. Consequently, the total excavation cost can be taken as lOxy = (— ) (x 2 y). This suggests that the 

S10/ft 2 



unit cost of excavation is ^—^ — where x is the length of a side of the square base of the tank in feet. For the least 

" 18 The total cost of the least expensive tank is 



expensive tank, the unit cost for the excavation is 1 / ft t - ■■-—■ 

$3375, which is the sum of $2625 for fabrication and $750 for the excavation. 



sn.fir 

ft3 



11. The area of the printing is (y — 4)(x — 8) = 50. 

Consequently, y = (j^g) + 4. The area of the paper is 
A(x) = x (^ + 4) , where 8 < x. Then 



A'(x) 



50 



+ 4 -x 



[o^Wj 



4(x-8) 2 -400 







vx-8 ' 7 \(x-syj (x-8) 2 

=> the critical points are —2 and 18, but —2 is not in the 
domain. Thus A"(18) > => atx = 18 we have 
a minimum. Therefore the dimensions 18 by 9 inches 
minimize the amount of paper. 



4 

2 2 

oo 

i 

X 

y-4 

4 



12. The volume of the cone is V = h 7rr 2 h, where r = x = i/9 — y 2 and h = y + 3 (from the figure in the text). 
Thus, V(y) = f (9 - y 2 ) (y + 3) = § (27 + 9y - 3y 2 - y 3 ) => V'(y) = § (9 - 6y - 3y 2 ) = tt(1 - y)(3 + y). 
The critical points are —3 and 1, but —3 is not in the domain. Thus V"(l) = | (—6 — 6(1)) < =£- at y = 1 
we have a maximum volume of V(l) = | (8)(4) = =y£ cubic units. 



Copyright (c) 2006 Pearson Education 




248 Chapter 4 Applications of Derivatives 



13. The area of the triangle is A(0) 



ab sin 6 



Solving A'(0) = 



ab cos 



where < < ir. 
= f. Since A"(0) 



A" (|) < 0, there is a maximum at 




14. A volume V = 7rTh = 1000 



h 



1000 



The amount of 



material is the surface area given by the sides and bottom of 



the can 

dS _ 

dr 

are and 



2000 _ 
r 2 
10 

3/3 



= 27rrh + 7rr 2 
2ttt= 



2000 



TIT 3 - 1000 



7rr 2 , < r. Then 
= 0. The critical points 



but is not in the domain. Since 



d_s _ 4goo _|_ 2tj- > o, we have a minimum surface area when 

r = 4t= cm an d h = i^r = 4t= cm - Comparing this result to 

the result found in Example 2, if we include both ends of the 
can, then we have a minimum surface area when the can is 
shorter-specifically, when the height of the can is the same as 
its diameter. 



15. With a volume of 1000 cm and V = 7rr 2 h, then h = 
A = 8r 2 + 27rrh = 8r 2 + 2S2Q _ Then A ' (r) = 16r 



The amount of aluminum used per can is 



2000 

r 2 







8r 3 -1000 



=>■ the critical points are and 5, 



40 



but r = results in no can. Since A (r) = 16 + ^tt > we have a minimum atr = 5 =>■ h = — and h:r = 8 



:tt. 



16. (a) The base measures 10 — 2x in. by 



15-2x 



in., so the volume formula is V(x) 



x(10-2x)(15-2x) 
2 



2x 3 - 25x 2 + 75x. 



(b) We require x > 0, 2x < 10, and 2x < 15. Combining these requirements, the domain is the interval (0, 5). 

V 

80- 

60 - 

40 

20 

-20 

(c) The maximum volume is approximately 66.02 in. 3 when x w 1.96 in. 
(1.9618739,66.019118) 





(d) V'(x) = 6x 2 — 50x + 75. The critical point occurs when V'(x) = 0, at x 



50± v /(-50) 2 -4(6)(75) _ 50 



±V700 



2(6) — 12 

= ' J - v , that is, x w 1.96 or x w 6.37. We discard the larger value because it is not in the domain. Since 
V"(x) = 12x — 50, which is negative when x w 1.96 , the critical point corresponds to the maximum volume. The 

25-5^ 



maximum volume occurs when x 



6 



1.96, which comfimrs the result in (c). 



17. (a) The" sides" of the suitcase will measure 24 — 2x in. by 18 — 2x in. and will be 2x in. apart, so the volume formula is 

V(x) = 2x(24 - 2x)(18 - 2x) = 8x 3 - 168x 2 + 862x. 



Copyright (c) 2006 Pearson Education 




Dt 



Section 4.5 Applied Optimization Problems 249 

(b) We require x > 0, 2x < 18, and 2x < 24. Combining these requirements, the domain is the interval (0, 9). 




-400 



(c) The maximum volume is approximately 1309.95 in. 3 when x « 3.39 in. 



1600 1 (3.3944503,1309.9547) 

1200 




-400 



(d) V'(x) = 24x 2 - 336x + 864 = 24(x 2 - 14x + 36). The critical point is at x 



14±y / (-14) 2 -4(l)(36) i 4±> /52 

2(1) 2 

= 7 ± y 13, that is, x w 3.39 or x « 10.61. We discard the larger value because it is not in the domain. Since 
V"(x) = 24(2x — 14) which is negative when x w 3.39, the critical point corresponds to the maximum volume. The 
maximum value occurs at x = 7 — y 13 w 3.39, which confirms the results in (c). 

(e) 8x 3 - 168x 2 + 862x = 1120 => 8(x 3 - 21x 2 + 108x - 140) = => 8(x - 2)(x - 5)(x - 14) = 0. Since 14 is not in 
the fomain, the possible values of x are x = 2 in. or x = 5 in. 

(f) The dimensions of the resulting box are 2x in., (24 — 2x) in., and (18 — 2x). Each of these measurements must be 
positive, so that gives the domain of (0, 9). 



18. If the upper right vertex of the rectangle is located at (x, 4 cos 0.5 x) for < x < it, then the rectangle has width 2x and 
height 4 cos 0.5x, so the area is A(x) = 8x cos 0.5x. Solving A'(x) = graphically for < x < n, we find that 
x w 2.214. Evaluating 2x and 4 cos 0.5x for x ss 2.214, the dimensions of the rectangle are approximately 4.43 (width) by 
1.79 (height), and the maximum area is approximately 7.923. 



19. Let the radius of the cylinder be r cm, < r < 10. Then the height is 2y 100 — r 2 and the volume is 



V(r) = 271-rVlOO-r 2 cm 3 . Then, V'(r) = 2m 2 



x/100-r 2 



(-2r) + (27^100 - r 2 ) (2r) 



-2m 3 + 47ri-(100-r 2 ) 
\/l00-r 2 



27tt(200 - 3f 2 ) 
VlOO-r 2 



The critical point for < r < 10 occurs at r : 



10 



. Since V'(r) > for 



< r < 10a/ I and V'(r) < for 10W | < r < 10, the critical point corresponds to the maximum volume. The 



dimensions are r = 10 



.16 cm and h 



20 



11.55 cm, and the volume is 



4000?r 
3\/3 



2418.40 cm 3 



20. 



(a) From the diagram we have 4x + I = 108 and V = x 2 L 
The volume of the box is V(x) = x 2 (108 — 4x), where 
< x < 27. Then 

V'(x) = 216x - 12x 2 = 12x(18 - x) = 
=>• the critical points are and 18, but x = results in 
no box. Since V"(x) = 216 - 24x < at x = 18 we 
have a maximum. The dimensions of the box are 
18 x 18 x 36 in. 



Copyright (c) 2006 Pearson Education 




250 Chapter 4 Applications of Derivatives 



(b) In terms of length, V(f) = x 2 £ 



;^) 2 £. The graph 
indicates that the maximum volume occurs near I = 36, 
which is consistent with the result of part (a). 




21. (a) From the diagram we have 3h + 2w 
V = h 2 w => V(h) = h 2 (54 - \ h) 



108 and 
54h 2 - | h 3 . 



Then V'(h) = 108h - § h 2 = f h(24 - h) = 
=>- h = or h = 24, but h = results in no box. Since 
V"(h) = 108 - 9h < at h = 24, we have a maximum 
volume at h = 24 and w = 54 - \ h = 18. 



(b) 







10 15 20 25 30 35 



22. From the diagram the perimeter is P = 2r + 2h + 7rr, 
where r is the radius of the semicircle and h is the 
height of the rectangle. The amount of light transmitted 
proportional to 

A = 2rh + \ tit 2 = r(P - 2r - ?rr) + \m 2 



rP 



4 

2r 2 - \m 



2 Then ^ 

dr 



P - 4r - i Trr = 



2P 

8 + 3- 
2r _ 



2h 



4P 

8 + 3- 



2ttP 
8 + 3tt 



(4 + tt)P 

8 + 3- 



Therefore, £ 

' h 

most light since 



^— gives the proportions that admit the 



4 
di-2 



|tt<0. 




2r 



23. 



j3 — M , where h is the height of the cylinder and r is the radius 



The fixed volume is V = 7rr 2 h + § 7rr 3 =>■ h 

of the hemisphere. To minimize the cost we must minimize surface area of the cylinder added to twice the 

surface area of the hemisphere. Thus, we minimize C = 27rrh + 47rr 2 = 27rr (^ — If) + 47rr 2 = 2 j- 

=> V 



| Trr 2 . 



Then f -- 

4VV3 



f + 1 i*r: 



fTrr 3 



1^) . From the volume equation, h = ^ — y 



2-3i/3.yi/3 _ 3i/3. 2 .4.yi/3 _ 2-3i/3. v i/3 



^1/3.32/3 3-2-7rV3 3-2-tt1/3 

dimensions do minimize the cost. 



:f) V3 - Since § 



4V 
i-3 



7T > 0, these 



24. The volume of the trough is maximized when the area of the cross section is maximized. From the diagram 



the area of the cross section is A(9) = cos 8 + sin 8 cos 8, < 8 < f . Then A' (8) = —sin 8 



cos 2 8 — sin 2 ( 



= - (2 sin 2 6i + sin 8 - 1) 
sin 8 ^ - 1 when < 8 < 
there is a maximum. 



-(2 sin 8 - l)(sin + 1) so A'(6>) = 
. Also, A'(6>) > for < 



sin 1 



or sin 1 



I because 



< § and A'(0) < for | < < | . Therefore, at 



Copyright (c) 2006 Pearson Education 




Section 4.5 Applied Optimization Problems 25 1 



25. (a) From the diagram we have: AP = x, RA = y L — x 2 , 
PB = 8.5 - x, CH = DR=11-RA=11 \/h - x 2 
QB = y/x 2 - (8.5 - x) 2 , HQ = 11 -CH-QB 

= 11 - [ll - VL - x 2 + v/x 2 - (8.5 - x) 2 ] 

= VL-x 2 - v/x 2 - (8.5 - x) 2 , RQ 2 = RH 2 + HQ 2 

= (8.5) 2 + (Vl-x 2 - v/x 2 - (8.5 - x) 2 ) . It 




follows that RP 2 = PQ 2 + RQ 2 



L 2 = x 2 + (Vl 2 - x 2 - v/x 2 - (x - 8.5) 2 ) 2 + (8.5) 2 



L 2 - x 2 - ly/l? - x 2 v/17x-(8.5) 2 + 17x - (8.5) 2 + (8.5) 2 



L 2 = x 

72„2 _ A (1 'I v-'i ; Hv ,',' ^2, I 2 



=> 17 2 x 2 = 4 (L 2 - x 2 ) (17x - (8.5) 2 ) => L 2 



x 



2 



17¥ 



(b) Iff(x) 



4 [17x-(8.5) 2 ] 
17x 3 

is minimized, then L 2 is minimized. Now f'(x) 



Hx- 1 
17x- (8.5) 2 



4x 3 
4x- 17 



2X- 1 
2X-8.5' 



4x-17 



4x 2 (8x-51) 
(4x- 17) 2 



f '(x) < when x < f 



and f ' (x) > when x > 5± . Thus L 2 is minimized when x 



51 



8 



(c) When x 



'-,thenL« 11.0 in. 



L 
35 
30 
25 
20 
15 




^V*' 5 6 — 7 — 6 S lb x 



26. (a) From the figure in the text we have P = 2x + 2y => y = | — x. If P = 36, then y = 18 — x. When the 

cylinder is formed, x = 27rr =>■ r = ^- and h = y => h = 18 — x. The volume of the cylinder is V = 7rr 2 h 



V(x) 



18x 2 -x 3 



3x(12-x) 



4?r 



=>• x = or 12; but when x = 0, there is no cylinder. 



|J =>■ V (12) < =>■ there is a maximum at x = 12. The values of x = 12 cm and 



. . Solving V'(x) 
ThenV"(x) = \ (3 ^ ^ v "' 
y = 6 cm give the largest volume, 
(b) In this case V(x) = 7rx 2 (18 - x). Solving V'(x) = 37rx(12 - x) = =>• x = or 12; but x = would result in 
no cylinder. Then V"(x) = 67r(6 — x) =4> V"(12) < => there is a maximum at x = 12. The values of 
x = 12 cm and y = 6 cm give the largest volume. 



27. Note that h 2 



3 and so r = a/3 - h 2 . Then the volume is given by V = § r 2 h = |(3 - h 2 )h = Trh - f h 3 for 



< h < \/3, and 



so 



dV 



it — 7rr 



7r(l — r 2 ). The critical point (for h > 0) occurs at h = 1. Since ^ > for 



< h < 1, and ^ < for 1< h < 



3, the critical point corresponds to the maximum volume. The cone of greatest 
volume has radius y 2 m, height lm, and volume ^ m 3 . 



28. (a) f(x) = x 2 
(b) f(x) = x 2 



f'(x) = x~ 2 (2x 3 - a) , so that f'(x) = when x = 2 implies a = 16 
f"(x) = 2x~ 3 (x 3 + a) , so that f"(x) = when x = 1 implies a = -1 



29. If f(x) = x 2 + I , then f '(x) = 2x - ax~ 2 and f "(x) = 2 + 2ax~ 3 . The critical points are and 3 ^/| , but x ^ 0. 
= 6>0 => atx= 3 ,/1j there is a local minimum. However, no local maximum exists for any a. 



Now f 



11 ti 



30. If f(x) = x 3 + ax 2 + bx, then f'(x) = 3x 2 + 2ax + b and f"(x) = 

(a) A local maximum at x = — 1 and local minimum at x = 3 = 
27 + 6a + b = =>• a = -3 and b = -9. 

(b) A local minimum at x = 4 and a point of inflection at x = 1 



6x + 2a. 

, f'(-l) = 0andf'(3) = : 

=> f (4) = and f"(l) = 



3 - 2a + b = and 
48 + 8a + b = 



Cfiojt (c| 1 Pearson link, k, publishing as Pearson Addison-Wesle 



252 Chapter 4 Applications of Derivatives 

and 6 + 2a = => a = -3 and b = -24. 

31. (a) s(t) = -16t 2 + 96t + 112 =4> v(t) = s'(t) = -32t + 96. At t = 0, the velocity is v(0) = 96 ft/sec. 

(b) The maximum height ocurs when v(t) = 0, when t = 3. The maximum height is s(3) = 256 ft and it occurs at t = 3 
sec. 

(c) Note that s(t) = -16t 2 + 96t + 112 = -16(t + l)(t - 7), so s = at t = -1 or t = 7. Choosing the positive value 
of t, the velocity when s = is v(7) = —128 ft/sec. 



32. 



6 mi ■ 




Village 



Jane 



Let x be the distance from the point on the shoreline nearest Jane's boat to the point where she lands her boat. Then she 
needs to row y 4 + x 2 mi at 2 mph and walk 6 — x mi at 5 mph. The total amount of time to reach the village is 

- i. Solving f'(x) = 0, we 



f(x) 



have: 



yT 



+ tp hours (0 < x < 6). Then f'(x) 



2 2\/4 + x 2 



(2x) 



2y/T- 



2^/T- 



5x = 2^4 + x 2 => 25x 2 = 4(4 + x 2 



21x 2 



16 



± -$—. We discard the negative 

\/21 



value of x because it is not in the domain. Checking the endpoints and critical point, we have f(0) = 2.2, 

f( -j— j m 2.12, and f(6) « 3.16. Jane should land her boat -4— w 0.87 miles donw the shoreline from the point 

nearest her boat. 



33. 



216 



and L(x) 

2 



h 2 + (x + 27) 2 



,2 + (x+27) 2 is 




x + 27 "^ " 

■ + ^Y + (x + 27)^ when x > 0. Note that L(x) is 

minimized when f(x) = (8 + -''' 
minimized. If f'(x) = 0, then 
2(8+ 2 ^)(- 2 J#)+2(x + 27)=0 
=>■ (x + 27) (l - i|n) = => x = -27 (not acceptable 
since distance is never negative or x = 12. ThenL(12) = \/2197 « 46.87 ft. 

34. (a) From the diagram we have d 2 = 4r 2 — w 2 . The strength of the beam is S = kwd 2 = kw (4r 2 — w 2 ) . When 
r = 6, then S = 144kw - kw 3 . Also, S'(w) = 144k - 3kw 2 = 3k (48 - w 2 ) so S'(w) = =$> w = ± 4^ ; 
S" (4^3 ) < and — 4y 3 is not acceptable. Therefore S (4a/ 3 j is the maximum strength. The dimensions 

of the strongest beam are 4 y 3 by 4 \J 6 inches, 
(b) (c) 





4 6 8 10 12 

Both graphs indicate the same maximum value and are consistent with each other. Changing k does not 
change the dimensions that give the strongest beam (i.e., do not change the values of w and d that produce 
the strongest beam). 



Copyright (c) 2006 Pearson Education 




Section 4.5 Applied Optimization Problems 253 



35. (a) From the situation we have w 2 = 144 — d 2 . The stiffness of the beam is S = kwd 3 = kd 3 ( 144 — d 



2\l/2 



where < d < 12. Also, S'(d) 



4kd 2 (108-d 2 ) 



critical points at 0, 12, and by 3. Both d = and 



(b) 



%/144-d 2 

d = 12 cause S = 0. The maximum occurs at d = 6 y 3. The dimensions are 6 by 6y 3 inches. 

(c) 





"J i 6 S 10 12" 

Both graphs indicate the same maximum value and are consistent with each other. The changing of k has 
no effect. 



=> sin t = sin t cos | + sin | cos t =>• sin t = | sin 



36. (a) Si = S2 => sin t = sin (t + |) 
==> t = f or y" 

(b) The distance between the particles is s(t) = |si — S2I = sin t— sin (t+ f ) = \ 



t + Y- cos t ==> tan t = \/3 
sin t — v 3 cos t 



s'(t) 



t-V3 



sin t — v 3 cos t cos 



t+v^s 



critical times and endpoints 



the 



2 sin t — \/3 cos t 

are 0, f , f , f , l -f, 2tt; then s(0) = ^ , s (§) = 0, s (f ) = 1, s (y") = 0, s (±±£) = 1, s(2tt) = ^ 
greatest distance between the particles is 1. 

I sin t— v 3 cos t J ( cos t+ y 3 sin 1 1 . 

(c) Since s (t) = . '—^p 1 we can conclude that at t = f and 3?, s (t) has cusps and 

2 sin t - n/3 cos t } J 

the distance between the particles is changing the fastest near these points. 



37. (a) s=10cos(7rt) =>- v = — 107r sin(7rt) =>■ speed = |107r sin (7rt)| = 107r | sin (7rt)| => the maximum speed is 

107T w 31.42 cm/sec since the maximum value of |sin (7rt)| is 1; the cart is moving the fastest at t = 0.5 sec, 
1.5 sec, 2.5 sec and 3.5 sec when |sin (?rt)| is 1. At these times the distance is s = 10 cos (|) = cm and 
a = -107r 2 cos(7rt) =>• |a| = 107r 2 | cos (7rt) | => |a| = cm/sec 2 
(b) |a| = 10-7T 2 |cos (7rt)| is greatest at t = 0.0 sec, 1.0 sec, 2.0 sec, 3.0 sec and 4.0 sec, and at these times the 
magnitude of the cart's position is |s| = 10 cm from the rest position and the speed is cm/sec. 

38. (a) 2 sin t = sin 2t =>• 2 sin t — 2 sin t cos t = =>• (2 sin t)(l — cos t) = =>• t = kir where k is a positive 

integer 

(b) The vertical distance between the masses is s(t) = |sj — s 2 | = ((sj — s 2 ) ) = ((sin 2t — 2 sin t) 2 ) 
=> s'(t) = (i) ((sin 2t - 2 sin t) 2 )~ 1/2 (2)(sin 2t - 2 sin t)(2 cos 2t - 2 cos t) 

_ 2(cos2t-cost)(sin2t-2sint) _ 4(2 cos t + l)(cos t - l)(sin t)(cos t - 1 ) critical times at 

|sin 2t— 2 sin t| |sin 2t — 2 sin t| 

0, f , 7T, f , 2tt; then s(0) = 0, s (f ) = |sin (f ) - 2 sin (f ) | = ^ , s(tt) = 0, s (f ) 



sin | -> ) — 2 sin (^) | 



: s f) 



3V3 



3V3 



T-, s(27r) = => the greatest distance is ^^ at t = % and y 



2\l/2 



39. (a) s = v/(12- 12t) 2 + (8t) 2 = ((12 - 12t) 2 + 64t 

(b) | = i ((12 - 12t) 2 + 64t 2 r 1/2 [2(12 - 12t)(-12) + 128t] 



208t-144 
x /(12-12t) 2 + 64t 2 



ds I 

dt I i=o 



1 2 knots and 



ds I 



8 knots 



Cop # (c) 1 Pearson Ediratio^, k, publishing as Pearson Addison-Wesle 



254 Chapter 4 Applications of Derivatives 



(c) The graph indicates that the ships did not see 
each other because s(t) > 5 for all values of t. 



(d) The graph supports the conclusions in parts (b) 
and (c). 



(e) lim g 

t — » 00 at 



i: m (208t- 144)2 

t" 5o 144(1 - 1)2 + 64t2 



lim 



208 - 



| t^"So 144 (l-l)%-64 V 144 + 64 

which equals the square root of the sums of the squares of the individual speeds 

40. The distance OT + TB is minimized when OB is 
a straight line. Hence Za = Z/3 =£> 9\ = Qi- 














B 


A 










b 


* 


-\ 


\8, 


si 


p 






J 


JT J 




• 













41. If v = kax — kx 2 , then v' = ka — 2kx and v" = —2k, so v' = 
v" ( l) = — 2k < 0. The maximum value of v is ^f- . 



Atx 



there is a maximum since 



42. (a) According to the graph, y'(0) = 0. 

(b) According to the graph, y'(— L) = 0. 

(c) y(0) = 0, so d = 0. Now y'(x) = 3ax 2 + 2bx + c, so y'(0) = implies that c = 0. There fore, y(x) = ax 3 + bx 2 and 
y'(x) = 3ax 2 + 2bx. Then y(-L) = -aL 3 + bL 2 = H and y'(-L) = 3aL 2 - 2bL = 0, so we have two linear 
equations in two unknowns a and b. The second equation gives b = t=. Substituting into the first equation, we have 



-aL 3 + as 



3aL 3 



H, or ^y- = H, so a = 2 ft. Therefore, b = 3ft and the equation for y is 



y(x)=2fty+3fty,ory(x)=H[2(^) 3 + 3(^) 2 ]. 



43. The profit is p = nx - nc = n(x - c) = [a(x - c)" 1 + b(100 - x)] (x - c) = a + b(100 - x)(x - c) 

= a + (be + 100b)x - lOObc - bx 2 . Then p'(x) = be + 100b - 2bx and p"(x) = -2b. Solving p'(x) = =^> 
x = | + 50. At x = | + 50 there is a maximum profit since p"(x) = —2b < for all x. 

44. Let x represent the number of people over 50. The profit is p(x) = (50 + x)(200 - 2x) - 32(50 + x) - 6000 

= -2x 2 + 68x + 2400. Then p'(x) = -4x + 68 and p" = -4. Solving p'(x) = => x = 17. At x = 17 there is a 
maximum since p"(17) < 0. It would take 67 people to maximize the profit. 



Copyright (c) 2006 Pearson Education 




45. (a) A(q) = kmq : + cm + | q, where q > =^> A'(q) 



critical points are -J^ , 0, and J ^ , but only W 2 ^ is in the domain. Then A" L/ ^ ] > =• ai 



Section 4.5 Applied Optimization Problems 

kmq- 2 + | = ^^ and A"(q) = 2kmq- 3 . The 

km\ 



255 



there is a minimum average weekly cost. 



(b) A(q) 



(k+bq)m 

q 



q = kmq l + bm + cm + | q, where q > => A'(q) = at q 



2 ^s as in (a). 



Also A"(q) = 2kmq 3 > so the most economical quantity to order is still q 
the average weekly cost. 



which minimizes 



46. We start with c(x) = the cost of producing x items, x > 0, and 



c(x) 



to be differentiable. If the average cost can be minimized, it will be at a production level at which dx 

x c'(x) — c(x) 



the average cost of producing x items, assumed 

(by the quotient rule) =>• x c'(x) — c(x) = (multiply both sides by x 2 ) =>• c'(x) = — where 
c'(x) is the marginal cost. This concludes the proof. (Note: The theorem does not assure a production level that will give a 
minimum cost, but rather, it indicates where to look to see if there is one. Find the production levels where the average cost 
equals the marginal cost, then check to see if any of them give a mimimum.) 



47. The profit p(x) = r(x) - c(x) = 6x - (x 3 - 6x 2 + 15x) 



6x 2 — 9x, where x > 0. Then 



p'(x) = — 3x 2 + 12x — 9 = — 3(x — 3)(x — 1) and p' (x) = — 6x + 12. The critical points are 1 and 3. Thus 
p"(l) = 6>0 => atx=l there is a local minimum, and p"(3) = —6 < =>• at x = 3 there is a local maximum. 
But p(3) = => the best you can do is break even. 

The average cost of producing x items is c(x) = ^ = x 2 - 20x + 20, 000 => c'(x) = 2x - 20 = =>• x = 10, the 
only critical value. The average cost is c(10) = $19, 900 per item is a minimum cost because c"(10) = 2 > 0. 



49. (a) The artisan should order px units of material in order to have enough until the next delivery. 

(b) The average number of units in storage until the next delivery is ^ an d so the cost of storing then is s(^) per 
day, and the total cost for x days is (^)sx. When added to the delivery cost, the total cost for delivery and storage 
for each cycle is: cost per cycle = d + ^sx. 



(c) The average cost per day for storage and delivery of materials is: average cost per day = 
To minimize the average cost per day, set the derivative equal to zero, j- ( d(x)~ + ^x 



(d+f* 2 ) d 

X X 

= - d(x) " 



+ ^=0 



=> x = ± J |j. Only the positive root makes sense in this context so that x* 
minimum, check the second derivative ^ ( — d(x)~ + y 



— . To verify that x* gives a 



3 >0 



a minimum. 



The amount to deliver is px* = 
(d) The line and the hyperbola intersect when 2 = E£x. Solving for x gives Xj ntersect j on = ± » / — . For x > 0, 

Xintersection = \ if = x*. From this result, the average cost per day is minimized when the average daily cost of 
delivery is equal to the average daily cost of storage. 



50. Average Cost 



. c« 



_di 

dx- 1 



(**) 



x=100 



I"' in 
100 3 



^ + 96 + 4X 1 / 2 ^> ^ h&\ = -^ + 2X" 1 / 2 = ^> x = 100. Check for a minimum: 
100" 3/2 = 0.003 > => a minimum at x = 100. At a production level of 100, 000 units, 



the average cost will be minimized at $156 per unit. 



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256 Chapter 4 Applications of Derivatives 



51. We have || = CM - M 2 . Solving |§ = C - 2M = 
maximum. 



M 



2 ■ A1SO, dM 3 



-2<0=>atM=§ there is ; 



52. (a) If v = crgr 2 — cr 3 , then v' = 2cror — 3cr 2 = cr (2tq — 3r) and v" = 2crg — 6cr = 2c (ro — 3r) . The solution of 



/ = is r = or ^ , but is not in the domain. Also, v' > for r < Sfc and v' < for r > ^a 



at 



r = ?p there is a maximum. 

(b) The graph confirms the findings in (a), 
v 




oTT O oT3 oTi o.s 



53. If x > 0, then (x — l) 2 > =>• x 2 + 1 > 2x =>• ^ti > 2. In particular if a, b, c and d are positive integers, 
*en (*±i) (Sfi) (*±i) (^i) > 16. 



54. (a) f(x) 



f'(x) 



( a 2+x 2 ) 1/2 -x 2 (a 2 +x 2 )- 1/2 



.,_ _|_ V Z — V Z 



v / a 2 + x 2 ~" " w (a 2 + x 2 ) 

f(x) is an increasing function of x 



(a 2 + x 2 ) 3/2 (a 2 + x 2 ) 3/2 



>0 



(b) g(x) 



^/b 2 + (d - x) 2 
- (b 2 + (d - x) 2 ) + (d - x) 2 



,,, _ - (b 2 + (d - x) 2 ) 1/2 + (d - x) 2 (b 2 + (d - X) 2 )" 1 -' 2 
■ W — b 2 + (d - x) 2 



(b 2 + (d - x) 2 ) 3/2 (b 2 + (d - x) 2 ) 3/2 



< => g(x) is a decreasing function of x 



(c) Since ci, C2 > 0, the derivative $- is an increasing function of x (from part (a)) minus a decreasing 



dx 

function of x (from part (b)): $■ = — f(x) — — g(x) 



dx Ci 



C2 



d 2 t _ 1 fl 
dx 2 c^ 



f '00 - f g'(x) > since f'(x) > and 



g'(x) < => j- is an increasing function of x. 



55. At x = c, the tangents to the curves are parallel. Justification: The vertical distance between the curves is 
D(x) = f(x) — g(x), so D'(x) = f'(x) — g'(x). The maximum value of D will occur at a point c where D' = 0. At 
such a point, f'(c) — g'(c) = 0, or f'(c) = g'(c). 

56. (a) f(x) = 3 + 4 cos x + cos 2x is a periodic function with period 2tt 

(b) No, f(x) = 3 + 4 cos x + cos 2x = 3 + 4 cos x + (2 cos 2 x - 1) = 2 (1 + 2 cos x + cos 2 x) = 2(1 + cos x) 2 > 
=>• f(x) is never negative 

57. (a) If y = cot x — y 2 esc x where < x < it, then y' = (esc x) ( y 2 cot x — esc x j . Solving y' = 

=>• cos x = -j- => x = I . For < x < | we have y' > 0, and y' < when | < x < n. Therefore, at x = | 
there is a maximum value of y = — 1 . 



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(b) 



Section 4.5 Applied Optimization Problems 257 



"0T5 I its 1 2T5 5" : 




The graph confirms the findings in (a). 



58. (a) If y = tan x + 3 cot x where < x < | , then y' = sec 2 x — 3 esc 2 x. Solving y' = 



tan x 



±^3 



=>• x = ± |, but — I is not in the domain. Also, y" = 2 sec 2 x tan x + 3 esc 2 x cot x > for all < x < | . 
Therefore at x = | there is a minimum value of y = 2\/3. 



(b) 



80 



y = tanx+3cotx 



0.25 0.5 0.75 1 1.25 1.5 

The graph confirms the findings in (a). 



59. (a) The square of the distance is D(x) 



x- 



+ (s/x + 0) 2 = x 2 - 2x + |, so D'(x) = 2x - 2 and the critical 



point occurs at x = 1. Since D'(x) < for x < 1 and D'(x) > for x > 1, the critical point corresponds to the 
minimum distance. The minimum distance is ^/D(l) = %r- . 



(b) 







D(x) = x 2 - 2x + ^ 
1 4 


2.5 






2 






1.5 




>/^^--**> = ^c 


1 

0.5 


1 


1 -* 1 1 *■! 



0.5 1 1.5 



The minimum distance is from the point ( | , 0) to the point ( 1 , 1 ) on the graph of y 
value x = 1 where D(x), the distance squared, has its minimum value. 

60. (a) Calculus Method: 

The square of the distance from the point ( 1, \J 3 j to ( x, \J 16 — x 2 j is given by 

D(x) = (x-1) 2 - 



^/x, and this occurs at the 



16 



2x + 1 + 16 - x 2 - 2^48 - 3x 2 + 3 = - 2x + 20 - 2 a/48 - 3x 2 . 



ThenD'(x) 



2- 



\/48 - 3x 2 



(_6x) = - 2 + 



0x 



\/48 - 3x 2 



Solving D'(x) = we have: 6x = 2\/48 - 3x 2 



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258 Chapter 4 Applications of Derivatives 



36x 2 = 4(48 - 3x 2 ) => 9x 2 



3x 2 



12x 2 



± 2. We discard x 



-2 as an extraneous solution, 



leaving x = 2. Since D'(x) < for —4 < x < 2 and D'(x) > for 2 < x < 4, the critical point corresponds to the 
minimum distance. The minimum distance is ^/D(2) = 2. 
Geometry Method: 

The semicircle is centered at the origin and has radius 4. The distance from the origin to I 1 , y3 ) is 



l 2 + I a/3 ) = 2. The shortest distance from the point to the semicircle is the distance along the radius 
containing the point I 1, \J 3 ) . That distance is 4 — 2 = 2. 



(b) 




The minimum distance is from the point ( 1, y 3 j to the point ( 2, 2y 3 j on the graph of y = \J 16 — x 2 , and this 
occurs at the value x = 2 where D(x), the distance squared, has its minimum value. 



61. (a) The base radius of the cone is r = 2 ™ v ~ an d so me height is h = v a 2 — r 2 = y a- 



2n 



. Therefore, 



V(x) = fr 2 h=f(^^) 



' 27ra - x \ 2 
v 2tt ) ■ 



3' " 3 ' 

(b) To simplify the calculations, we shall consider the volume as a function of r: volume = f(r) = ^r 2 y a 2 — r 2 , where 



< r < a. f ' 

2a 2 r - 3r 3 



'(r) = l^tVtfTTpi) = | L . i ^= ? (-2r) + (v^^)(2r) 



-r 3 + 2r(a 2 - r 2 ) 

\/a 2 — r 2 



^ 



h= \f 



a z 



7rr(2a 2 - 3r 2 ) 
3\/a 2 - r 2 



,2 2a 2 



The critical point occurs when r 2 = ^-, which gives r = a-t " 



a\/6 



Then 



^-. Using r : 



n/o 



and h 



n/3 



we may now find the values of r and h 



for the given values of a. 



When a = 4 
When a = 5 
When a = 6 
When a = 8 



3 ' 3 ' 

5y6 i _ 5y3 , 



3 ' " 3 ' 

r= 21/6^= 2a/3; 

r _ sy^e h - 5\/3- 
1 — 3 , 11 — 3 , 



(c) Since r = -|- and h = -^-, the relationship is jj = v 2. 



62. (a) Let Xo represent the fixed value of x at the point P, so that P has the coordinates (xo, a), and let m = f'(xo) be the 
slope of the line RT. Then the equation of the line RT is y = m(x — x ) + a. The y-intercept of this line is 
m(0 — xo) + a = a — mxo, and the x-intercept is the solution of m(x — Xo) + a = 0, or x = m *°~ a . Let O designate 
the origin. Then 
(Area of triangle RST) 

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Section 4.5 Applied Optimization Problems 259 



2(Area of triangle ORT) 

2 • |(x-intercept of line RT)(y-intercept of line RT) 

(a - mx ) 

mxp — a ^ / mxp — a ^ 



9 1 f mn-o - 
A 2V m 



/ mx - a \2 
V m / 



Substituting x for Xq, f'(x) for m, and f(x) for a, we have A(x) = — f'(x) 



M 

f'(x) 



(b) The domain is the open interval (0, 10). To graph, let yi = f(x) = 5 + 5i/l — j^, y2 = f'(x) = NDER(yi), and 
y3 = A(x) = — y2 1 x — — J . The graph of the area function y3 = A(x) is shown below. 

A(x) 
500 
400 
300 
200 
100 




— < 1 f— 



2 4 6 8 10 

The vertical asymptotes at x = and x = 10 correspond to horizontal or vertical tangent lines, which do not form 
triangles. 
(c) Using our expression for the y-intercept of the tangent line, the height of the triangle is 



a — mx 



f(x) -f'(x) -x = 5+ WlOO-x 2 - 



5+Wl00-x 2 + - 7 4 i 



2^100- 



2 2^100 - x 2 2 

We may use graphing methods or the analytic method in part (d) to find that the minimum value of A(x) occurs at 
x w 8.66. Substituting this value into the expression above, the height of the triangle is 15. This is 3 times the 
y-coordinate of the center of the ellipse, 
(d) Part (a) remains unchanged. Assuming C > B, the domain is (0, C). To graph, note that 

1 



f(x)=B + B«/l-£ =B + Pv / C 2 -x 2 andf'(x) 



B 



A(x) = -f'(x) 



M 

f'(x) 



Bx 



cVc 2 - x 2 



B + ^v'C 2 -x2 



C 2\/C 2 -X 2 

2 



(-2x) 



-Bx 



cVa'- 



■ . Therefore we have 



C\ c 2 -x 2 



Bx 



(BC + B\/C2~ 



C 2 -x 2 



2 



C\/C 2 -x 2 



Bx 



BCx\/C 2 - x 2 

1 
BCxx/C 2 - x 2 



Bx 2 + (BC + Ba/C^-x 2 ) (JC 2 - x 2 ) 



BCx\/C 2 - x 2 



Bx 2 + BCa/C 2 - x 2 + B(C 2 - x 2 



Bc(c+ v^-x 2 ) 



BC(C + \/C 2 - x 2 J 
x\/C 2 - x 2 



(xy^^ 2 -)(2)(c + yc 2 ^ 2 -)(-^ ? )-(c +v ^^ 2 -) (x-^+y/c^q)) 
a v xj - c^ • - x2(c2 - x2) 



Bc(c + v^-x 2 
x 2 (C 2 - x 2 ) 



-2x 2 - (c + a/c 2 - x 2 



x/c 2 " 



a/C 2 - x 2 ) 



BC(C + s/&-iA 

x 2 (C 2 - x 2 ) 



Cx 2 



C^C^ 



BC(C + \/C 2 -x 2 ) / 

x 2 (C 2 - x 2 ) (TcJ 

BC 2 (c+\/C 2 -x 2 ) , ,- 

^-^ W i(2x 2 -C 2 -C^ 



-2x 2 + ^SL^ _ cVc 2 - x 2 + x 2 -(C 2 - x 2 ) 



(c + \/C 2 - x 2 ) 



n/c 2 



C 



') 



BC 



X 2( C 2 _ X 2)V 2 



Cx 2 - C(C 2 



c 2 V^ 



To find the critical points for < x < C, we solve: 2x 2 - C 2 = cVc 2 - x 2 => 4x 4 - 4C 2 x 2 + C 4 = C 4 - C 2 x 2 
=>• 4x 4 — 3C 2 x 2 = =>• x 2 (4x 2 — 3C 2 ) = 0. The minimum value of A(x) for < x < C occurs at the critical point 

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260 Chapter 4 Applications of Derivatives 

x = — j- , or x 2 = 2£L. The corresponding triangle height is 

a — mx = f(x) — f (x) ■ x 



+ 


y&- 


-X 2 


+ 


Bx 2 




Cv / C 2- 


3C2 
4 








+ 


B(3| 


"J 


+ 


IVc 2 - 


X 2 


C^- 


3C2 

4 


+ 


§(§) + 


4 








+ 


B , 3B 

2 ' 2 











= 3B 
This shows that the traingle has minimum arrea when its height is 3B. 



4.6 INDETERMINATE FORMS AND LHOPITALS RULE 



1. l'Hopital: lim 



x-2 _ 1 



7 or lim 



x-2 



lim 



x-2 



lim 



x " 2 x 2 - 4 2x | x=2 4 U1 x ^" 2 x 2 - 4 ^^ (x - 2) (x + 2) x ^' 2 x + 2 4 



2. l'Hopital: lim 



sin 5x 5 cos 5x 



lx=0 



5 or lim 



sin 5x 



5 lim 
5x^0 



sin 5x 
5x 



5-1=5 



3. l'Hopital: lim ^ 

r x — » oo ' 



lim 



10x-3 



lim It = I or lim 



5x 2 -3x 



5x 2 - 3x _ ^^ 

x — » oo 7x 2 + l x^i'oo 14x x^oo 14 7 """"x^+'oo 7x 2 +l x "^»"bo 7- 



lim 



lim 



3x- 



4. l'Hopital: ^lin^ 4x3 x _ x _ 3 - ^^ T25P^- : n ul x "^j 4x 3 - x -3 - x ^\ ( x -i)(4x 2 +4x+3) 



or lim 



lim 



(x-l)(x 2 + x+l) 



lim 

x^ 1 



(x 2 + x+l) _ 3 



(4x 2 +4x + 3) 11 



5. l'Hopital: lim 



x^O 



lim 

x^O 



2x 



lim 

x^O 



2 2 



lim "" x , 

x ^0 x 2 (l + cosx) 



lim 

x^O 



' sin x \ ( sin x \ ( 1 



x / V x / V 1 +cos x/ 



or lim 

x^O 

1 

2 



lim 

x^O 



(1-cosx) /l + cosx'i 



x- V 1 + cos X / 



6. l'Hopital: lim 

1 Y — > r 



2x 2 + 3x _ 
x ^"oo x 3 + x + 1 x — > oo 3x 2 + 1 x 



lim #H = lim #■ = or lim M+3* 



oo 6x 



lim 



\ -ox 3 + x+l x^ool + -L + 4j 1 



7. lim ^ = lim ?^^ = 



8. lim 2*=z = lim -?- = 4r = -2 

X^tt/2 cosx 0^tt/2 - 8mx - 1 

9. lim 5»L| = lim ^ = 4 = 1 



10. lim '- sin -* = lim - cosx = lim sinx 

x^tt/2 1 + cos2x x ^ 7r / 2 -2sm2x x _> ^ 



11. lim 

X — > 7r/4 



sin x — cos x 



lim 

X — > 7r/4 



cos x + sin x V 2 ■ y 2 

1 — 2 ' 2 



12. lim 



lim 



x^tt/3 x ~3 x^tt/3 : 



-sin x V 3 

2 



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i n A 



Section 4.6 Indeterminate Forms and L'Hopital's Rule 261 



13. lim — ( x — 5)tanx= lim 



x — ^ sin x 



X — > 7r/2 



x^ tt/2 



-. (S — x) cos x + sin x(— 1) 

lim ^ — : 

x^tt/2 - Slnx 



14. lim 



lim 



lim 



4y^ 



4-0 



x'^To x + ^v 7 * x _^ 1 + jV x"A"o 2^A+7 " 2-0+7 



15. lim 

x^ 1 



2x 2 -(3x + l) % /x + 2 



| im 2x^-3^ xV 2 + 2 = Hm 4X -i*f-^ = .J 



x^ 1 



x^ 1 



y/^Ts-3 _ ,, m K* 2 + 5 )" 1/2 C 2 *) _ ,;,„ 



16. lim V V J ,~ = lim 

x ^ 2 x2 - 4 x ^ 2 



2x 



lim — = - - 

x^2 2Vx 2 + 5 6 



17. lim 

x^O 



^/a(a + x) -a 



lim — ^ = -4- 

x ->0 2Va 2 + ax 2\/a 2 



5, where a > 0. 



18. Um iO(!HL^i) = Hm lO(cost-l) = lO^sint) = ^ost = ^HM = _ 5 

i ' t _» n 3t 2 t-tO 6t t-tO 6 6 3 



t-»0 



t->0 



x(cos X — 1) 



-xsin x + cqs x — 1 



19. lim X ^ C0SX ~ I; = lim 

x ^q sin x — x x > Q COS X — 1 



lim 

x _, -sin x 



xcos x — 2sin x \im xcos x + 2sin x 

~ x -> sin x 



lim 

x -> cos x 



xsin x + 3 cos x 3 



20 H m sin(a + h)-sina _ j^ cos(a + h) -cos a _ g 



h->0 



h->0 



21. lim — — =-^ = lim ^j — - = an lim r n 1 = an, where n is a positive integer. 

r— >l rl r— >!' r — > 1 



22. lim 

x^0+ 



23. lim 

X — > OO 



fl - 1 ) = hm (i^) = f raop,,al'srule\ = jj m A _ /^ . 1 = 

V x \/x/ x _> o+ V x / y does not apply y x ^ g+ V V I x 

- ^^) = x^oo (* - ^^) (xTTlS) 

_ _ 1 / l'Hopital's rule \ 
2 \ is unnecessary y 



lim 



x- - ( x^ + x ] 



lim 



X — ► OC X + <Jx 2 + X X — > OO x . Vx'^ + x 

x+^2^ 



lim — =± 

x ^°° i + ,/i 



24. lim xtan(i) = lim 

X — » OO V X/ X-KXJ 



lim 

X — > OO 



I 



lim sec 

X — » OO 



2(t\ 



sec^O = 1 



25. lim 



3x-5 



lim 



26. lim 

x^0 



x^±oo 2x ' x + 2 x ^ ± oo 
sin7x _ r;™ 7cos(7x) 



tan llx 



lim 



■1 _ 7 



q llsec 2 (llx) 111 11 



27. lim ^^ 

x — > oo v x + 1 



lim 



9x+l 



X — > oo x+ 1 



lim I = y/9 = 3 

x — > oo 1 v 



28. lim -)£- 

x _> 0+ Vsinx 



lim 512JL VI 

t^0+ x 



1 = 1 



29. lim 



lim 



x^tt/2" ranx x^tt/2- 



<• 1 ) I cos X > 
V cos x / V sin x / 



lim -4- = 1 

x ^ w /2~ Slnx 



30. lim 



lim 



x ^ + cscx x^0+ (Jrz) x^0+ 



lim cos x = 1 



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262 Chapter 4 Applications of Derivatives 

3 1 . Part (b) is correct because part (a) is neither in the § nor — form and so l'Hopital's rule may not be used. 



32. Answers may vary. 

(a) f(x) = 3x + l;g(x) = x 



lim 



f(x) 



lim 



3x + l 



lim 



X — > OO gW X — > OG X x^ool 

,2 



(b) f(x) = x+l;g(x)=x 2 



lim -^r = lim 



x+l 



lim 



x'-^J'oo g(x) x^+'oc x 2 x^+'oo 2x 

,2. r 



(c) f(x) = x 2 ;g(x)=x+l 



lim 



KM 



lim -4- 



lim 



X — > oo g(x) x->oo x+l x ^ oo 1 



OO 



33. If f(x) is to be continuous at x = 0, then lim f(x) = f(0) => c = f(0) = lim 9 *- 3s 3 in3 * = lim 

x — > x — > -' x x — > 



9 — 9 cos 3x 



lim 

x^O 



27 sin 3x 
30x 



lim 

x^O 



31 cos 3x 27 



30 ~~ 10 



34. (a) For x ^ 0, f'(x) = -f (x + 2) = 1 and g'(x) = -f (x + 1) = 1. Therefore, lim ^1 = 1 = 1, while lim ®& 

._ x + 2 _ + 2 _ 9 
x+l 0+1 z " 

(b) This does not contradict l'Hopital's rule because neither f nor g is differentiable at x = 

(as evidenced by the fact that neither is continuous at x = 0), so l'Hopital's rule does not apply. 



35. The graph indicates a limit near — 1 . The limit leads to the 
indeterminate form § : lim — — p^ — 

„ . 1 x— l 



lim 

x^ l 



x^ 1 

2x 2_ 3x 3/2, x l/2 + 2 _ 
X- 1 

I - 1=1 _ _ 
~~ 1 ~~ 



4x-|x 1/2 -ix-V2 

lim ' — ; — ' 

X->1 1 



2x 2 -(3x + l )Vx+2 




36. (a) 



-0.5 



20 40 60 80 100 



= X-VX 2 +X 



(b) The limit leads to the indeterminate form oo — oo 



lim 

X — > oo 



( x — V x 2 + x J = lim ( x — v x 2 + x 



; + \/x 2 +x 
c + yx 2 +x 



lim 

X — » oo 



/ x 2 -(x 2 + x) \ _ 
V x+\/x 2 + x/ x 



lim 



oo x+ y/x 2 + > 



lim 



1 + 1/1 + 



I l + Vl + 



x^0 



37. Graphing f(x) = - — Sjjjp- on th window [— 1, 1] by [—0.5, 1] it appears that lim f(x) = 0. However, we see that if we let 
u = x 6 , then lim f(x) = lim i^f^ = lim ^ = 1L,. , 

x ^ v ' u^0 u u ^ 2u u^O 2 



lim 



38. (a) We seek c in (-2, 0) so that ( M = ^llV^ = g^f = -\. Since f '(c) = 1 and g'(c) = 2c we have that £ = -\ 



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Section 4.7 Newton's Method 263 



(b) We seek c in any open interval (a, b) so that -^ 



f'(c) _ f(b) - f(a) _ b-a b-a _J_ . J_ _ _J_ 

(c) g(b)-g(a) b 2 -a 2 (b-a)(b + a) b + a ^ 2c b + a 



(c) We seek c in (0, 3) so that -4 



f'(c) _ f(3)-f(0) -3-0 

■ g(3)-g(0) " 9-0 



2c 



■\ => 3c 2 + 2c - 12 = => c 



(Note that c 



-l-v/37 



is not in the given interval (0, 3).) 



b + a 



-1+ -y/37 



39. (a) By similar triangles, f§ = g§ where E is the point on AB such that CE ± AB : 



















c . 


B(l,8) 










\e 






v^-~^\ 


/e 










(x,0) 1 


o 


D 7a(1,0) 





1 — x 1 — COS 

9 ~ 9 -sin 

(b) lim (1 - x) 



Thus ^—q^ = 4 — ^SMr, since the coordinates of C are (cos 0, sin 6). Hence, 1 — x = -4 — ^r^- 

6* 9 — sin 9 ' V ' / ' 9 — sin 

"1-cosfl) |._ 9sin9 + l-cos9 _ i;„, 9 cos g + sin 9 + sin 9 _ i- 9cos9 + 2sin< 



8^0 



lim 



lim 



lim 

0^0 



n > q 9 — sin 9 rt ^q 1 — cos 9 

sin 9) + cos 9 + 2cos 9 _ ,. -9sin9 + 3cos9 _ + 3 

cos B n . Q cos 1 



lim 

8^0 



sin 9 



lim 

8^0 



sin 9 



(c) We have that lim [(1 - x) - (1 - cos 0)1 = lim 

9 — > oo 9 — » oc 



- (1 -COS0) 



lim (1 — cos 9) 

9 — » oo 



As — ► oo, (1 — cos 8) oscillates between and 2, and so it is bounded. Since lim 

8 — > oo 



9 - sin 9 

— J ) - 1—1 = 



- 1 



lim (1 — cos 6) 

8 — > oo 

approaches 0. 



- 1 



0. Geometrically, this means that as 6 — » oo, the distance between points P and D 



40. Throughout this problem note that r 2 = y 2 + 1, r > y and that both r — > oo and y — » oo as 



(a) lim r — y = 

(b) n lim r 2 — y 2 



lim -i- = 

-tt/2 r +y 

lim 1 = 1 

9 -» tt/2 



6»^tt/2 
(c) We have that r 3 - y 3 = (r - y)(r 2 + ry + y 2 ) 



r + ry + y - y + y-y + y 
r + y ■* r 



3y : 



f- = 3y?. 



Since lim 3y 

9 -> tt/2 



lim 3sin # • y = oo we have that lim r 3 — y 



-» tt/2 



6»^tt/2 



4.7 NEWTON'S METHOD 



i- y 



x 2 
*2 



1 =>. y' = 2x + 1 

t+I-l 7 

> x 2 = i 



2x„ + l ' X 



9 _ 4-2-1 

^ A i 1 



2 _ 4+6-9 _ 
? 12+9 

-1.66667 



21 



13 

21 



1 => Xi 

61905; x = 



l+i-i 

2+1 

' X : = 



1-1-1 

-2+1 



3x + 1 =4> y' = 3x 2 + 3 =4> x n+ 



x 2 



-l+i 



J+3 



1 + 1 



90 



29 
90 



xg+3x„+l . 

3x2+3 ' X 

-0.32222 



=> Xj = 



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264 Chapter 4 Applications of Derivatives 



3. y = x 4 + x - 3 => y' = 4x 3 + 1 =4> x n+1 = x n 

1296 , 6 o 

625 ~t~5 J 



^±^^;x =l => x 1 = l i±l=2 



Y _ 6 

X 2 - 5 



1296+750-1875 



-2 => x 2 = -2 - 



+ 1 5 4320+625 

16-2-3 __ 2+ l 



-32+1 



4x3+1 

_ 6 _ J71_ _ 5763 r 
5 4945 4945 " 

-§} rs -1.64516 



4+1 



1.16542; x = -1 => Xj = -1 - 



1-1-3 



2x n -x2+l 



4. y = 2x - x 2 + 1 => y' = 2 - 2x => x n+1 = x„ - ^-2x7' ; x o = ° ^ x i = ° 

~'~' +1 -_l+i=_i« -.41667; x = 2 => xi = 2 4 ~ 4+1 



0-0+1 
2-0 



X 2 2 2+1 

5 _ 20-25+4 _ 5 _ 1 



-12 



2 12 



2 ' 12 12 

§ « 2.41667 



2-4 



1 
2 

X 2 



5-f+l 
2-5 



5. y = x 4 - 2 =>• y' = 4x 3 =^ x 

5 _ 113 _ 2500-113 _ 2387 
~~ 4 2000 ~~ 2000 ~~ 2000 



\ 4 - 
4x2 



" -;x = l =^ x x = 1- V 



*2 



5 _ 625-512 
4 2000 



1.1935 



6. From Exercise 5, x n 



x 4 — 2 . ,1-2 

x » - ip- ; x o = -i => xi = -l - y 



X 2 = - 7 



5 _ 625-512 
4 -2000 



113 
2000 



1.1935 



7. f(x ) = and f (x ) ^ 



f(x„ 



v„+l -a." f77x"J gives X! = x => x 2 = x 

the approximations in Newton's method will be the root of f(x) = 0. 



x n = Xq for all n > 0. That is, all of 



8. It does matter. If you start too far away from x = | , the calculated values may approach some other root. 
Starting with Xq = —0.5, for instance, leads to x = — | as the root, not x = | . 



9. Ifx = h>0 
/h 



Xi =x 



f(xo) 
f'(x ) 



h 



f(h) 
f'(h) 



= h 

ifx = 



-h<0 



h - (v/h) (2>/h) = -h; 

^ x i= x o-^ = -h- 
= -h+ (Vh) h-yfh) =h. 



f(-h) 
f'(-h) 




~{i ,x>0 
■4^x,x<0 



10. f(x) = x 1 / 3 => f'(x) 



-2/3 



,1/3 



Q): 



-2/3 



= — 2x n ; Xo = 1 =>• Xi = —2, x 2 = 4, X3 = —8, and 
X4 = 16 and so forth. Since |x n | = 2|x„_,| we may conclude 

that n — > 00 =>• IxJ — > 00. 




11. i) is equivalent to solving x 3 — 3x — 1 = 0. 
ii) is equivalent to solving x 3 — 3x — 1 = 0. 
iii) is equivalent to solving x 3 — 3x — 1 = 0. 
iv) is equivalent to solving x 3 — 3x — 1 = 0. 
All four equations are equivalent. 



12. f(x) = x - 1 - 0.5 sin x =^> f'(x) =1-0.5 cos x 
X! = 1.49870 



1 — 1 — 0.5 sin x. 
1 — 0.5 cos x„ 



- ;if x = 1.5, then 



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13. For Xq = —0.3, the procedure converges to the root —0.32218535. 
(a) 



Plotl Plots PlotJ 

\ylBx A 3+3x+l 
■*.y2BnDer(yl ix.x) 



r. I V I INSf I DELf ISELCTt 



(b) 



'.3-*x 



(C) 



x-al/y2-»x 

-.322324152194 
-.322185360292 
-.322135354626 
-.322135354626 



(d) Values for x will vary. One possible choice is Xq = 0.1. 



x-yl^y2-*x 



.1 



. 329372795537 
-.322200595043 
-.322135354698 
-.322135354626 



(e) Values for x will vary. 



Section 4.7 Newton's Method 265 



14. (a) f(x) = x 3 - 3x - 1 =>• f'(x) = 3x 2 - 3 => x n+I 

and -0.34730 
(b) The estimated solutions of x 3 — 3x — 1 = are 
-1.53209,-0.34730, 1.87939. 



x; - 3x„ - l 

3*?-3 



the two negative zeros are —1.53209 



(c) The estimated x-values where 

g(x) = 0.25x 4 — 1 .5x 2 — x + 5 has horizontal tangents 
are the roots of g'(x) = x 3 — 3x — 1, and these are 
-1.53209,-0.34730, 1.87939. 




-2/ -1 



ffx) = x 3 -3x-l 



g(x) = 0.25x 4 -1.5x 2 -x + 5 




15. f(x) = tan x - 2x => f'(x) = sec 2 x - 2 => x n+1 = x„ 
=> x 2 = 1.155327774 => x 16 = x 17 = 1.165561185 



tan(x„)-2x„ 
sec 2 (x n ) 



; x = 1 => xi = 12920445 



16. f(x) = x 4 - 2x 3 - x 2 - 2x + 2 =4> f'(x) = 4x 3 - 6x 2 - 2x - 2 => x„ 
ifx = 0.5, then x 4 = 0.630115396; if x = 2.5, then x 4 = 2.57327196 



X n ZX n X n ZX n + L 



n n n 



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266 Chapter 4 Applications of Derivatives 

17. (a) The graph of f(x) = sin 3x — 0.99 + x 2 in the window 
— 2 < x < 2, —2 < y < 3 suggests three roots. 
However, when you zoom in on the x-axis near x = 1.2, 
you can see that the graph lies above the axis there. 
There are only two roots, one near x = — 1 , the other 
near x = 0.4. 
(b) f(x) = sin 3x - 0.99 + x 2 =>■ f ' (x) = 3 cos 3x + 2x 

and the solutions 



sin(3x„) - 0.99+x 2 



-v a, 1+1 - ^ n 3 cos(3Xn ) + 2x„ 

are approximately 0.35003501505249 and 
-1.0261731615301 



y - sln(3x) - 0.99 + x 




18. (a) Yes, three times as indicted by the 
graphs 
(b) f(x) = cos 3x - x =>■ f ' (x) 
= —3 sin 3x — 1 => x n+ . 



cos(3x n )-x n 



; at 



11 —3 sin (3x n ) — 1 

approximately —0.979367, 
-0.887726, and 0.39004 we have 
cos 3x = x 





19. f(x) = 2x 4 - 4x 2 + 1 =>• f'(x) = 8x 3 



_ 2x;j-4x 2 +l .- 
X » 8x3 _ 8x . II X 



-2, thenx G = -1.30656296; if 



Xo = —0.5, then X3 = —0.5411961; the roots are approximately ± 0.5411961 and ± 1.30656296 because f(x) is 
an even function. 



20. f(x) = tan x => f '(x) = sec 2 x 
approximate -it to be 3.14159. 



tan (x n ) 
sec 2 (x„) 



;x 



xi = 3.13971 =* x 2 = 3.14159 and we 



21. From the graph we let Xo = 0.5 and f(x) = cos x — 2x 

=* x "+' = x » " -li^J-^ => x i = - 45063 

=^> X2 = .45018 =^> at x « 0.45 we have cos x = 2x. 



y 


3 




2 


/ y = 2x 


-3 yf -1 J 


1 \2 3 


— ^ /l 




y = cos x / 




/ " 2 




' -3 





22. From the graph we let x = —0.7 and f(x) = cos x + x 

=► X »+> = X » " T^SxT =► * = - 73944 

=^ X2 = —.73908 => at x ss —0.74 we have cos x = — x. 





3 


' 




\y = -x 2 


~">. y = cos x 


-3 


-/ -1 

-1 

-2 


\1 N2 3 




-3 





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Section 4.7 Newton's Method 267 



by the Intermediate Value Theorem the equation 

x^ + 2x n -4 



23. If f(x) = x 3 + 2x - 4, then f(l) = - 1 < and f(2) = 8 > 

x 3 + 2x — 4 = has a solution between 1 and 2. Consequently, f'(x) = 3x 2 + 2 and x n+1 -a,- - - , , ,— . 

Thenx =1 =>• x x = 1.2 => x 2 = 1.17975 =>• x 3 = 1.179509 =>• x 4 = 1.1795090 =>• the root is approximately 
1.17951. 

24. We wish to solve 8x 4 - 14x 3 - 9x 2 + 1 lx - 1 = 0. Let f(x) = 8x 4 - 14x 3 - 9x 2 + llx - 1, then 

.3 ^-.2 io_ , ,, 8x4 -14x3- 9x2 + llx„-l 



f (x) = 32x 3 - 42x 2 - 18x + 1 1 



32x- ! -42x 2 -18x n + ll 



x 


approximation of corresponding root 


-1.0 


-0.976823589 


0.1 


0.100363332 


0.6 


0.642746671 


2.0 


1.983713587 



25. f(x) = 4x 4 - 4x 2 



f'(x) = 16x 3 - 8x 



f(x.) 

POO 



Ix? 



- . Iterations are performed using the 



procedure in problem 13 in this section. 

(a) For Xo = 2 or Xo = —0.8, x; — > — 1 as i gets large. 

(b) For Xo = —0.5 or Xo = 0.25, x; — > as i gets large. 

(c) For Xo = 0.8 or Xo = 2, x; — > 1 as i gets large. 

(d) (If your calculator has a CAS, put it in exact mode, otherwise approximate the radicals with a decimal value.) 

Forx = -^orx = -^, 

/21 



Xo 



^orx 



Newton's method does not converge. The values of x; alternate between 
as l increases. 



26. (a) The distance can be represented by 



D(x) = J(x - 2) 2 + (x 2 + i) 2 , where x > 0. The 

distance D(x) is minimized when 

f(x) = (x - 2) 2 + (x 2 + |) 2 is minimized. If 

f(x) = (x - 2) 2 

f'(x) = 4 (x 3 + x - 1) and f"(x) = 4 (3x 2 + 1) > 0. 

Now f'(x) = 



(x 2 + i) 2 ,then 



x- 1 =0 =>■ x(x 2 + 1) = 1 



X 2+l 




(b) Letg(x) 



X 2+l 



(x 2 + 1)- 



?'(x) 



(x 2 + 1) z (2x) - 1 



-2x 



W 



;x = l 



(x 2 +l) 2 

X4 = 0.68233 to five decimal places. 



27. f(x) = (x - 1) 4U => f'(x) = 40(x - l) 3y => x n 



x„ — 



40(x n -l) 3 



39x„ + 1 
40 



With x = 2, our computer 



jave x 87 = x 88 = x 89 



x 200 = 1 ■ 1 105 1 , coming within 0. 1 105 1 of the root x = 1 . 



"n (K ~ 1) 



28. f(x) = 4x 4 - 4x 2 => f'(x) = 16x 3 - 8x = 8x (2x 2 - 1) =4> x n+1 = x„ - ^ _ ^ ; if x = .65, then 

X12 « -.000004, if x = .7, then x 12 = -1.000004; if x = .8, then x 6 = 1.000000. NOTE: ^ « .654654 



29. f(x) = x 3 + 3.6x 2 - 36.4 =4> f'(x) = 3x 2 + 7.2x =>■ x„ 



xj + 3.6x 2 - 36.4 



; x = 2 => xi = 2.5303 



»+'- A " 3x2 + 7.2x n 

x 2 = 2.45418225 =>• x 3 = 2.45238021 =>• x 4 = 2.45237921 which is 2.45 to two decimal places. Recall that 

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268 Chapter 4 Applications of Derivatives 

x = 10 4 [H 3 0+] => [H 3 0+] = (x) (10~ 4 ) = (2.45) (1(T 4 ) = 0.000245 
30. Newton's method yields the following: 



the initial value 


2 


i 


V^ + i 


the approached value 


1 


-5.5593H 


-29.5815 - 17.0789i 



4.8 ANTIDERIVATIVES 



1. 


(a) 


X 2 


2. 


(a) 


3x 2 


3. 


(a) 


x- 3 


4. 


(a) 


-x- 2 


5. 


(a) 


X 


6. 


(a) 


1 

X 2 


7. 


(a) 


v 7 ^ 


8. 


(a) 


x 4/3 


9. 


(a) 


x 2/3 


10. 


(a) 


x l/2 


11. 


(a) 


COS (7TX) 


12. 


(a) 


sin (7rx) 


13. 


(a) 


tan x 


14. 


(a) 


—cot X 


15. 


(a) 


—esc X 


16. 


(a) 


sec x 



17. J(x+ l)dx= f +x + C 
19. J(3t 2 + |)dt = t 3 + f +C 



(V) 


3 


(b) 


X s 
8 


(b) 


X" 3 

3 


(b) 


X -2 1 X" 

4 ~r 3 


(b) 


^5 

X 


(b) 


-1 
4x2 


(b) 


v^ 


(b) 


1 Y 2/3 

2 X 


(b) 


x l/3 


(b) 


x -l/2 


(b) 


—3 cos x 


(b) 


sin(f) 


(b) 


2 tan (f) 


(b) 


cot(f) 


(b) 


j esc (5x) 


(b) 


| sec(3x) 



(C) y - X 2 + X 



(c) f - 3x 2 + 8x 
(c) - ^ + x 2 + 3x 



(C) V + T 

5 



(c) 2x 



(c) 



4 ^ 2x2 



(c) fv^ + 2^ 

( C ) 3 x 4/3 + 3 x 2/3 

(c) x" 1 / 3 



(c) x- 3 / 2 



(c) 



—COS (7TX) 



cos(3x) 



(c) (f)sin(f)+ 7 r l 

(c) - § tan (|) 

(c) x + 4cot(2x) 

(c) 2csc(f) 

(c) fsec(f) 



18. / (5 - 6x) dx = 5x - 3x 2 + C 
20. J U + 4t 3 )dt=£ + t 4 + C 



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Section 4.8 Antiderivatives 269 



21. J (2x 3 - 5x + 7) dx = i x 4 - § x 2 + 7x + C 22. J" (1 - x 2 - 3x 5 ) dx = x - | x 3 - i x G + C 



23- J(^-x 2 -i)dx=/(x- 2 -x 2 -i; 



dx 



-13 3 



x + C 



2x 2 



1_ X^ X 

x 3 3 



^ +C ==- + ^+X Z + C 



c = -^ + c 



24- / Q - Jr +2x) dx = / (i - 2x- 3 + 2x) dx = ± x - (^) 

25. fx- 1 /3 d x=^+C=|x 2 / 3 +C 26. f x - 5 / 4 dx=^ + 

27. / (^ + Vx) dx = / (x 1 / 2 + x 1 / 3 ) dx =^ + ^ + C=? x 3 / 2 + | x 4 / 3 + C 

28. / (^ + -^) dx = / (i x 1 / 2 + 2X- 1 / 2 ) dx = 1 (^) + 2 (*£) + C = | x 3 / 2 + 4X 1 / 2 + C 

29. /(8y-^)dy = /(8y-2y- 1 / 4 )dy=¥-2(f)+C = 4y 2 -fy 3 / 4 + C 



30. /(*-£) dy = /($-r^ 4 )dy = *y-(^)+C = $ + £ + C 



31. / 2x (1 - x- 3 ) dx = / (2x - 2x~ 2 ) dx = ^ - 2 fe) 



C = x 



2 i 2 



32 



x 2x 2 + *~ 



J x- 3 (x + 1) dx = J (x- 2 + x- 3 ) dx = £i + f il) + C 

33. /^dt = /(¥ + ¥)*=/(f 1/s +r' /!, )*=^+(^)+c = 2 > /t 

34. J^dt=/(i + ^)dt=/(4t- 3 + t-/ 2 )dt = 4(^) + (^) + C 



* +C 



t 2 3t 3/2 T «- 



35. J" -2 cos t dt = -2 sin t + C 

37. /7sin | d0 = -21 cos f + C 

39. J" -3 esc 2 x dx = 3 cot x + C 

41. j cscg 2 cotfl d0 = - \ csc 6 + C 

43. J (4 sec x tan x — 2 sec 2 x) dx = 4 sec x — 2 tan x + C 



36. J -5 sin t dt = 5 cos t + C 

38. J 3 cos 50 d0 = | sin 50 + C 

40. / - ^ dx = - ^p + C 

42. / § sec 6> tan d0 = | sec (9 + C 



44. J i (csc 2 x — csc x cot x) dx = — i cot x + | csc x + C 



45. J (sin 2x — csc 2 x) dx = — \ cos 2x + cot x + C 46. J (2 cos 2x — 3 sin 3x) dx = sin 2x + cos 3x + C 

47. / l±s&* dt = / (I + | cos 4t) dt = \ t + \ (^) + C = \ + ^ + C 

48. / Mf^ dt = / (i - i cos fit) dt = 1 1 - 1 (^S) + C = \ - ^ + C 

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270 Chapter 4 Applications of Derivatives 

49. / (1 + tan 2 0) 6.9 = / sec 2 Ad = tan + C 

50. / (2 + tan 2 9) d0 = f (1 + 1 + tan 2 0) d0 = / (1 + sec 2 0) d0 = + tan + C 

51. J cot 2 x dx = J (esc 2 x — 1) dx = — cot x — x + C 

52. f (1 - cot 2 x) dx = J (1 - (esc 2 x - 1)) dx = J" (2 - esc 2 x) dx = 2x + cot x + C 

53. / cos (tan + sec 0) d0 = / (sin + 1) d0 = -cos + + C 

54. f f cS . , d0 = f ( f cB . g ) (*Pf) d0 = f t-4-j, d0 = f ^j-j d0 = f sec 2 d0 = tan i 

J csc 6 — sin J V esc 6 — sin / V sin ) J 1 —sin 2 6 J cos 2 9 J 



55- ^( (Z ^+c) = 4 ^#^=(7x-2)3 



56- £ 



(_ i^x+^i + ^ = _ £_ (3» + 5)-'(3) j = (3x + 5) . 



57. £ (i tan(5x - 1) + C) = ± (sec 2 (5x - 1)) (5) = sec 2 (5x - 1) 



58. 



59. 



dx 



(_3 cot (i=i) + C) = -3 (-csc 2 (s=i)) (i) = csc 2 (s=i) 



dx V x 



i + c) = (-i)(-i)( X +ir 



(X+1)2 



uu - dx Vx+1 



c) 



(x+l)(l)-x(l) 

(X + 1)2 



(X+l)2 



sin x + y cos x = x sin x + y cos x 7^ x sin x 



61. (a) Wrong: £ (f sin x + c) = 

(b) Wrong: £ (— x cos x + C) = —cos x + x sin x ^ x sin x 

(c) Right: gj (— x cos x + sin x + C) = —cos x + x sin x + cos x = x sin x 

62. (a) Wrong: £ (^ + c) = 3 - J f^ (sec tan 0) = sec 3 tan ^ tan sec 2 

(b) Right: £ (i tan 2 + C) = 1 (2 tan 0) sec 2 = tan sec 2 

(c) Right: ^ (1 sec 2 + C) = \ (2 sec 0) sec tan = tan sec 2 

63. (a) Wrong: £ (2* + i£ + c) = 3(2x + 1)2(2) = 2(2x + l) 2 / (2x + l) 2 

(b) Wrong: £ ((2x + l) 3 + C) = 3(2x + 1) 2 (2) = 6(2x + l) 2 f 3(2x + l) 2 

(c) Right: £ ((2x + l) 3 + C) = 6(2x + l) 2 



64. (a) Wrong: £ (x 2 + x + cf' 2 = \ (x 2 + x + C) 1/L (2x + 1) = 2y/ ^ + c + ^2x + 1 
(b) Wrong: £ ((x 2 + x) 1/2 + c) = * (x 2 + x)~ 1/2 (2x + 1) = ^= ^ ^/2x + 1 



(c) Right: 



U (v/2x+l) 3 + CJ = £ Q (2x + I) 3 / 2 + C) = § (2x + l) 1 /2(2) = V / 2xTT 



65. Graph (b), because ^ = 2x => y = x 2 + C. Then y(l) = 4 => C = 3. 

66. Graph (b), because % = -x =>• y = - | x 2 + C. Then y(-l) = 1 => C = | 



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Section 4.8 Antiderivatives 271 



67. 



dy 



2x - 7 =>■ y = x 2 - 7x + C; at x = 2 and y = we have = 2 2 - 7(2) + C => C = 10 => y = x 2 - 7x + 10 



68. 



dy 



10 -x => y = lOx- \ +C;atx = Oandy = -1 we have -1 = 10(0)- \ +C =>• C = 1 



=$► y = 10x - \ - 1 
69. & - 1 



dx x 2 



x 1 + y+C;atx = 2 and y = 1 we have 1 



■U? 



+ c 



y = — x * + y — i or y 



1 , xf _ 1 
x + 2 2 



70. ^=9x 2 -4x + 5 => y = 3x 3 - 2x 2 + 5x + C; at x = -1 and y = we have = 3(-l) 3 - 2(-l) 2 + 5(-l) + C 
=> C = 10 =>• y = 3x 3 - 2x 2 + 5x + 10 



71. % = 3x- 2 / 3 => y=3^+C = 9;atx = 9X 1 / 3 + C; at x = -1 and y = -5 we have -5 = O(-l) 1 / 3 + C => C = 4 



d\ 



:ix' ■' 
l 

3 



y = 9X 1 / 3 + 4 



72. %- = -X- = \ x- 1 / 2 => y = x 1 / 2 + C; at x = 4 and y = we have = 4 1 / 2 + C^C = -2^y = x 1 / 2 - 2 



73. jjj = 1 + cos t => s = t + sin t + C; at t = and s = 4 we have 4=0+sinO + C => C = 4 =$■ s = t+sint+4 

74. gj = cos t + sin t => s = sin t — cos t + C; at t = it and s = 1 we have 1 = sin n — cos tt + C => C = 
=$> s = sin t — cos t 

75. g = -7r sin 7t6» =>■ r = cos(7T0) + C; at r = and = we have = cos(7r0) + C => C = -1 => r = cos(7T0) - 1 

76. % = cos tt0 => r = \ sin(7T0) + C; at r = 1 and = we have 1 = \ sin (vrO) + C=^C=l=^r=isin (tt0) + 1 



77. 



dv _ 1 



t tan t => v = \ sec t + C; at v = 1 and t = we have 1 = \ sec (0) + C => C = \ =4> v = \ sec t + \ 



78. 



dv 



,2 



J( - 8t + esc 2 1 => v = 4t 2 - cot t + C; at v = -7 and t = f we have -7 = 4 (| ) - cot (§) + C =>• C = — 7 - 7r 
=>• v = 4t 2 - cot t - 7 - 7T 2 



d^ 



Jy 



dy 



79. H = 2 - 6x ^> g = 2x - 3x 2 + Ci; at §- = 4 and x = we have 4 = 2(0) - 3(0) 2 + Ci => Ci = 4 

=4> g = 2x - 3x 2 + 4 =>• y = x 2 - x 3 + 4x + C 2 ; at y = 1 and x = we have 1 = 2 - 3 + 4(0) + C 2 => C 2 = 1 

=>■ y = x 2 - x 3 + 4x + 1 

80. g = => | = Ci; at ^ = 2 and x = we have d = 2 => g = 2 =4> y = 2x + C 2 ; at y = and x = we 
have = 2(0) + C 2 => C 2 = => y = 2x 

81. gf = 2 =2 r 3 ^ | = -r 2 + Ci;at| = landt= 1 we have 1 = -(1)~ 2 + Ci => Ci = 2 => g = -r 2 + 2 
=> r = r 1 + 2t + C 2 ; at r = 1 and t = 1 we have 1 = l -1 + 2(1) + C 2 => C 2 = -2 => r = t" 1 + 2t - 2 or 

r = ± + 2t - 2 



82. 



d 2 s _ 3t . ds _ 3t 2 



W-J => | = t+ C i; at l = 3andt = 4wehave3 =T? L + C i =* C i=° =* l = f => s=^+C 2 ;at 
s = 4 and t = 4 we have 4 = ^ + C 2 => C 2 = =^ s = ^ 



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272 Chapter 4 Applications of Derivatives 



83- g=6 



g = 6x + Ci; at = -8 and x = we have -8 = 6(0) + d => d = -8 — — 

g = 3x 2 - 8x + C 2 ; at g = and x = we have = 3(0) 2 - 8(0) + C 2 

3 _ Asr'2 i r„- ot u - s onH v — n ,1,0 havo ^ — n3 _ 4 cm 2 1 r". -^ r^ — ^ -■*. „ _ v 3 /i-»2 



dx2 

dy 



6x-8 



C 2 = =>- g = 3x 2 - 8x 



=4> y = X J - 4x 2 + C 3 ; at y = 5 and x = we have 5 = 3 - 4(0) 2 + C 3 => C 3 = 5 => y = x 3 - 4x 2 + 5 



84. ff = => ff = d; at ff = -2 and t = we have |f = -2 => f = -2t + C 2 ; at f = - \ and t = we 
have - \ = -2(0) + C 2 => C 2 = - | - - dfl - ^ 1 — fl - * 2 4 



d» 
dt 



-2t-i => 



r - 1 1 + C 3 ; at 9 = yjl and t = we have 



'I = -0 2 - | (0) + C 3 => C 3 = \/2 => 6» = -t 2 - i t + V2 



85. y (4) = -sin t + cos t =>- y'" = cos t + sin t + Ci; at y'" = 7 and t = we have 7 = cos(0) + sin(0) + Ci 

=> Ci = 6 => y'" = cos t + sin t + 6 => y" = sin t - cos t + 6t + C 2 ; at y" = - 1 and t = we have 
-1 = sin(0) - cos (0) + 6(0) + d =^ C 2 = =>• y" = sin t - cos t + 6t => y' = -cost- sint+ 3t 2 + C 3 ; 
at y' = - 1 and t = we have - 1 = -cos (0) - sin (0) + 3(0) 2 + C 3 => C 3 = =4> y' = -cos t - sin t + 3t 2 

=4> y = -sin t + cos t + t 3 + C 4 ; at y = and t = we have = -sin (0) + cos (0) + 3 + C 4 => C 4 = - 1 

=4> y = —sin t + cos t + t 3 — 1 

86. y' 4 ' = —cos x + 8 sin (2x) => y'" = — sin x — 4 cos (2x) + Ci; at y'" = and x = we have 

= -sin (0) - 4 cos (2(0)) + Ci =4> Ci = 4 =>■ y'" = -sin x - 4 cos (2x) + 4 => y" = cos x - 2 sin (2x) + 4x + C 2 ; 

at y" = 1 and x = we have 1 = cos (0) - 2 sin (2(0)) + 4(0) + C 2 => C 2 = => y" = cos x - 2 sin (2x) + 4x 
=> y' = sin x + cos (2x) + 2x 2 + C 3 ; at y' = 1 and x = we have 1 = sin (0) + cos (2(0)) + 2(0) 2 + C 3 => C 3 = 
=> y' = sin x + cos (2x) + 2x 2 => y — —cos x + i sin (2x) + § x 3 + C 4 ; at y = 3 and x = we have 

3 = -cos (0) + \ sin (2(0)) + § (0) 3 + C 4 =^C 4 = 4=^y = -cos x + \ sin (2x) + § x 3 + 4 

87. m = y' = 3Jx = 3x 1/2 => y = 2x 3 / 2 + C; at (9,4) we have 4 = 2(9) 3/2 + C => C = -50 =4> y = 2x 3 / 2 - 50 



I (a) g = 6x 



^ = 3x 2 + Ci; at y' = and x = we have = 3(0) 2 + d => d = 



l=3x 2 



=> y = x 3 + C 2 ; at y = 1 and x = we have C 2 = 1 => y = x 3 + 1 
(b) One, because any other possible function would differ from x 3 + 1 by a constant that must be zero because 
of the initial conditions 

89. g = 1- f x 1 / 3 => y = J(l - f-x 1 / 3 ) dx = x-x 4 / 3 + C; at (1,0.5) on the curve we have 0.5 = 1 - l 4 / 3 + C 

^ C = 0.5 => y = x- x 4 / 3 + \ 

90. g =x- 1 =>■ y = J(x- l)dx= ^ -x + C;at(-l,l)onthecurvewehave 1 = ^ -(-1) + C 

=* c = -i =► y=f- x -5 

91. g| = sin x — cos x =$■ y = J (sin x — cos x) dx = —cos x — sin x + C; at (— tt, —1) on the curve we have 
— 1 = — cos(— 7r) — sin(— 7r) + C => C = — 2 =4- y = — cos x — sin x — 2 

92. ^ = ^k- + tt sin 7rx = \ x~ 1/2 + 7r sin 7rx => y = / (| x~ 1/2 + sin 7rx) dx = x 1/2 - cos 7rx + C; at (1,2) on the 
curve we have 2 = l 1 / 2 — cos 7r(l) + C => C = =4- y= \fx — cos 7rx 



93. (a) f = 9.8t - 3 =4> s = 4.9t 2 - 3t + C; (i) at s = 5 and t = we have C = 5 => s = 4.9t 2 - 3t + 5; 

displacement = s(3) - s(l) = ((4.9)(9) - 9 + 5) - (4.9 - 3 + 5) = 33.2 units; (ii) at s = -2 and t = we have 
C = -2 =>■ s = 4.9t 2 - 3t - 2; displacement = s(3) - s(l) = ((4.9)(9) - 9 - 2) - (4.9 - 3 - 2) = 33.2 units; 
(iii) at s = So and t = we have C = so =>- s = 4.9t 2 — 3t + So; displacement = s(3) — s(l) 
= ((4.9)(9) - 9 + s ) - (4.9 - 3 + s ) = 33.2 units 

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Section 4.8 Antiderivatives 273 

(b) True. Given an antiderivative f(t) of the velocity function, we know that the body's position function is 

s = f(t) + C for some constant C. Therefore, the displacement from t = a to t = b is (f(b) + C) — (f(a) + C) 
= f(b) — f(a). Thus we can find the displacement from any antiderivative f as the numerical difference 
f(b) — f(a) without knowing the exact values of C and s. 



94. a(t) = v'(t) = 20 
= 1200 m/sec. 



v(t) = 20t + C; at (0, 0) we have C = => v(t) = 20t. When t = 60, then v(60) = 20(60) 



95. Step 1: p = -k => f = -kt + d; at f = 88 and t = we have d 
s = -k (| J + 88t + C 2 ; at s = and t = we have C 2 = => 



ds 
dt 



-kt + : 



Step 2: ^=0 
Step 3: 242 = 



= -kt + 88 



2 



> t = 

242 



88 
k 



(W , (88)1 
2k ' k 



242 



2k 



+ 88t 



16 



96. 



eft 
dt- 



ds 
dt 



/ -k dt = -kt + C; at | = 44 when t = we have 44 = -k(0) + C =>■ C = 44 



ds 
dt 



kt 2 



kt + 44 =$> s = - s- + 44t + Ci; at s = when t = we have 



kfO)' 



44(0) + Ci 



Ci =0 



f + 44t. Then f 



=>■ -kt + 44 = => t 



fands(f) 



^r + 44(f 



45 



968 , 1936 



45 



«6s 
k 



45 



968 



21.5 



ft 

sec 2 



97. (a) v = / a dt = j (l5t J / 2 - 3t- J / 2 ) dt = 10t 3 / 2 - 6I 1 / 2 + C; f (1) = 4 =^ 4= 10(1) 3 / 2 - 6a) 1 / 2 + C => C = 
=► v = 10t 3 / 2 - 6tV 2 
(b) s = / v dt = J*(l0t 3 / 2 - 6t J / 2 ) dt = 4t 5 / 2 - 4t 3 / 2 + C; s(l) = => = 4(1) 5 / 2 - 4(1) 3 / 2 + C => C = 
=>. s = 4t 5 / 2 - 4t 3 / 2 



eft 
dt 2 



-5.2 



i = -5.2t + Ci; at | = and t = we have Ci = 



ds 
dt 



-5.2t => s = -2.6t 2 + d; at s = 4 



and t = we have Co 



-2.6t 2 + 4. Then s = => = -2.6t 2 + 4 => t 



1.24 sec, since t > 



99. 



eft 
dt- 



=> | = J a dt = at + C; | = v when t = =^> C 



vn 



|=at + v 



v t + Ci;s = s 



when t = 



„ _ a(0) 2 
s — -^T 



v (0) + Ci =>• d = so 



f + v t + s 



100. The appropriate initial value problem is: Differential Equation: 



dt 2 



-g with Initial Conditions: 



Vq and 



s = Sq when t = 0. Thus, 



J-gdt 



? t + C i; |(0) = v => v = (-g)(0) + Ci =► Ci 



vo 



=► i = -8t + v . Thus s = /(-gt + v ) dt = - 1 gt 2 + v t + C 2 ; s(0) = s =-■ - | (g)(0) 2 + v (0) + C 2 =>• C 2 = s 
Thus s 



i gt 2 + v t + s . 



101. (a) Jf(x)dx= 1- v/x + d = -v/x + C (b) Jg(x) dx = x + 2 + Ci = x + C 

(c) J-f(x) dx = - (1 - y/x) + d = y/x+ C (d) J-g(x) dx = -(x + 2) + Ci = -x + C 

(e) / [f(x) + g(x)] dx = (1 - yfi) + (x + 2) + Ci = x - y/i + C 

(f) / [f(x) - g(x)] dx = (1 - y/x) - (x + 2) + Cx = -x - yfi. + C 

102. Yes. If F(x) and G(x) both solve the initial value problem on an interval I then they both have the same first 
derivative. Therefore, by Corollary 2 of the Mean Value Theorem there is a constant C such that 

F(x) = G(x) + C for all x. In particular, F(x ) = G(x ) + C, so C = F(x ) - G(x ) = 0. Hence F(x) = G(x) 
for all x. 



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274 Chapter 4 Applications of Derivatives 

103 — 106 Example CAS commands: 
Maple : 

with(student): 

f := x -> cos(x) A 2 + sin(x); 

ic := [x=Pi,y=l]; 

F := unapply( int( f(x), x ) + C, x ); 

eq := eval( y=F(x), ic ); 

solnC := solve( eq, {C} ); 

Y := unapply( eval( F(x), solnC ), x ); 

DEplot( diff(y(x),x) = f(x), y(x), x=0..2*Pi, [[y(Pi)=l]], 

color=black, linecolor=black, stepsize=0.05, title="Section 4.8 #103" ); 
Mathematica : (functions and values may vary) 

The following commands use the definite integral and the Fundamental Theorem of calculus to construct the solution 
of the initial value problems for exercises 103 - 105. 

Clear[x, y, yprime] 

yprime[x_] = Cos[x] 2 + Sin[x]; 

initxvalue = -it; inity value = 1; 

y[x_] = Integrate [yprime [t], {t, initxvalue, x}] + inityvalue 
If the solution satisfies the differential equation and initial condition, the following yield True 

yprime [x]==D[y[x],x] //Simplify 

y[initxvalue]==inity value 
Since exercise 106 is a second order differential equation, two integrations will be required. 

Clear[x, y, yprime] 

y2prime[x_] = 3 Exp[x/2] + 1; 

initxval = 0; inity val = 4; inityprimeval = — 1; 

yprime[x_] = Integrate[y2prime[t],{t, initxval, x}] + inityprimeval 

y[x_] = Integrate [yprime [t], {t, initxval, x}] + inityval 
Verify that y[x] solves the differential equation and initial condition and plot the solution (red) and its derivative (blue). 

y2prime[x]==D[y[x], {x, 2}]//Simplify 

y[initxval]==inityval 

yprime[initxval]==inityprimeval 

Plot[{y[x], yprime[x]}, {x, initxval - 3, initxval + 3}, PlotStyle -> {RGBColor[l,0,0], RGBColor[0,0,l]}] 

CHAPTER 4 PRACTICE EXERCISES 

1. No, since f(x) = x 3 + 2x + tan x => f'(x) = 3x 2 + 2 + sec 2 x > => f(x) is always increasing on its domain 



2. No, since g(x) = esc x + 2 cot x =^ g'(x) = —esc x cot x — 2 esc 2 x 
=>- g(x) is always decreasing on its domain 



sin 2 x sin 2 x 



-\- (cos x + 2) < 



3. No absolute minimum because lim (7 + x)(ll — 3x) 1,/3 = — oo. Nextf'(x) 
(11 -3x)^ -(7 + x)(ll -3xr 2 / 3 = (11 ( 7 1 3 l ) 3x y ) = ^$ 
Since f > if x < 1 and f < if x > 1, f(l) = 16 is the absolute maximum. 



,T1 -3x)2/3 " i i ; _ 3x )2/3 => x = 1 and x = T are critical points. 



4. f(x)=f±^ 



f'(x) 



a(x2 ' ( 1 ]:y +b) = " ( "^ + _ 2 y a) I f (3) = => -£ (9a + 6b + a) = => 5a + 3b = 0. 



We require also that f(3) = 1. Thus 1 



3a+b 



3a + b = 8. Solving both equations yields a = 6 and b = — 10. Now, 



f'(x) = 2(3x 2 _ 1)( * 3) so that f = | | +++ | +++ | . Thus f changes sign at x = 3 from 

(x 1} -1 1/3 1 3 

positive to negative so there is a local maximum at x = 3 which has a value f(3) = 1. 

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Chapter 4 Practice Exercises 275 

5. Yes, because at each point of [0, 1) except x = 0, the function's value is a local minimum value as well as a 
local maximum value. At x = the function's value, 0, is not a local minimum value because each open 
interval around x = on the x-axis contains points to the left of where f equals — 1. 

6. (a) The first derivative of the function f(x) = x 3 is zero at x = even though f has no local extreme value at 

x = 0. 
(b) Theorem 2 says only that if f is differentiable and f has a local extreme at x = c then f '(c) = 0. It does not 
assert the (false) reverse implication f'(c) = => f has a local extreme at x = c. 

7. No, because the interval < x < 1 fails to be closed. The Extreme Value Theorem says that if the function is 
continuous throughout a finite closed interval a < x < b then the existence of absolute extrema is guaranteed on 
that interval. 

8 . The absolute maximum is | — 1 1 = 1 and the absolute minimum is 1 1 = 0. This is not inconsistent with the Extreme Value 
Theorem for continuous functions, which says a continuous function on a closed interval attains its extreme values on that 
interval. The theorem says nothing about the behavior of a continuous function on an interval which is half open and half 
closed, such as [—1, 1), so there is nothing to contradict. 



9. (a) There appear to be local minima at x = —1.75 
and 1.8. Points of inflection are indicated at 
approximately x = and x = ± 1 . 




(b) f'(x) = x 7 - 3x 5 - 5x 4 + 15x 2 = x 2 (x 2 - 3) (x 3 - 5). The pattern y' 



indicates a local maximum at x = "y 5 and local minima at x = ± y 3 . 



(c) 




y = x 8 /8-x S /2-x 5 + 5x 3 



f(x) = x 8 /8-x 6 /2-x 5 + 5x 3 



I +++ I +++ I — | - 
-N/3 V^ ^ 



-+ 



1.72 1.74 1.76 1.78 



10. (a) The graph does not indicate any local 

extremum. Points of inflection are indicated at 



approximately x 



and x = 1. 



J 






. 600 




/ 


\ 400 




/ 


V"° 




/ 


" 2 \ 

-201 


/ 


2 


-40C 
-60C 


f(x) 


-8" 5 - Sx 



(b) f'(x) = x 7 - 2x 4 - 5 + $ = x~ 3 (x 3 - 2) (x 7 - 5) . The pattern f 



)(+- 





• Sx-4r+ 11 

X 



+++ indicates 



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276 Chapter 4 Applications of Derivatives 



a local maximum at x = y 5 and a local minimum at x = y2 . 



(c) 



1 .07437 - 




1 .2585 



1.2599 



11. (a) g(t) = sin 2 1 — 3t =$■ g'(t) = 2 sin t cos t — 3 = sin (2t) — 3 => g' < =>• g(t) is always falling and hence must 

decrease on every interval in its domain, 
(b) One, since sin 2 1 — 3t — 5 = and sin 2 1 — 3t = 5 have the same solutions: f(t) = sin 2 1 — 3t — 5 has the same 
derivative as g(t) in part (a) and is always decreasing with f(— 3) > and f(0) < 0. The Intermediate Value 
Theorem guarantees the continuous function f has a root in [—3, 0]. 

12. (a) y = tan 8 =4> -=| = sec 2 6 > =>■ y = tan 9 is always rising on its domain =>• y = tan 6 increases on every 

interval in its domain 
(b) The interval [J, 7r] is not in the tangent's domain because tan is undefined at 8 = | . Thus the tangent 
need not increase on this interval. 



13. (a) f(x) = x 4 + 2x 2 - 2 =>• f'(x) = 4x 3 + 4x. Since f(0) = -2 < 0, f(l) = 1 > and f'(x) > for < x < 1, we 
may conclude from the Intermediate Value Theorem that f(x) has exactly one solution when < x < 1 . 



(b) x 



2 _ -2 ± V4 + 8 



>0 => x 2 = a/3 - 1 and x > =4> x « v / - 7 320508076 « .8555996772 



pyrj > 0, for all x in the domain of ^y =^ y = ^tj is increasing in every interval in 



14. (a) y = ^ =► y' -- (x + 

its domain 
(b) y = x 3 + 2x => y' = 3x 2 + 2 > for all x => the graph of y = x 3 + 2x is always increasing and can never 
have a local maximum or minimum 

15. Let V(t) represent the volume of the water in the reservoir at time t, in minutes, let V(0) = ao be the initial 
amount and V(1440) = ao + (1400)(43,560)(7.48) gallons be the amount of water contained in the reservoir 
after the rain, where 24 hr = 1440 min. Assume that V(t) is continuous on [0, 1440] and differentiable on 
(0, 1440). The Mean Value Theorem says that for some t in (0, 1440) we have V'(t ) = ^'^Ip^ 

= *o + (1400X43^60X7.48) - ao = 456 160,320 gal = 316778 1/min . Therefore at t the reservoir's volume 

1440 1440 mm ' & u 

was increasing at a rate in excess of 225,000 gal/min. 

16. Yes, all differentiable functions g(x) having 3 as a derivative differ by only a constant. Consequently, the 
difference 3x — g(x) is a constant K because g'(x) = 3 = -jj (3x). Thus g(x) = 3x + K, the same form as F(x). 



17 Nq x = 1 ' ' ^ x ^ffi=--c fi«™ _ " ! 



x+l ' x+1 x+1 — x + 

d_ I x \ _ (x+l)-x(l) _ 1 &_ I -1 

dx\x+l/ (x+1) 2 (x+1) 2 dxlx+1 



j differs from — -j by the constant 1. Both functions have the same derivative 



18. f'(x) = g'(x) 



2x 



(x 2 +l) 2 



f(x) — g(x) = C for some constant C =£> the graphs differ by a vertical shift. 



19. The global minimum value of \ occurs at x = 2. 



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Chapter 4 Practice Exercises 277 

20. (a) The function is increasing on the intervals [—3, —2] and [1, 2]. 

(b) The function is decreasing on the intervals [—2, 0) and (0, 1]. 

(c) The local maximum values occur only at x = —2, and at x = 2; local minimum values occur at x = —3 and at x = 1 
provided f is continuous at x = 0. 



21. (a) t = 0, 6, 12 



(b) t = 3,9 



(c) 6 < t < 12 



(d) < t < 6, 12 < t < 14 




(b) at no time 



(c) < t < 4 



(d) 4<t< 



24. 




y = x 3 -3x Z + 3 



25. 



-x 3 + 6x 2 -9x + i 



1 




-1 


\ 1 /2 3 4 

-V \ 



26. 




y=g-(x 3 + 3x~-9x-2? 



27. 



28. 



/ 


i 








500 
400 




y 


(6, 432) 




300 










200 






/(4,256) \ 




100 










-2/^1 







2 4 6 8l 




/-100 












29. 



30. 



; 


y = x-3x 2 " 


i i. — , 


-3 


i. 9 _J2-— " 

(8, -4) 


27 




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278 Chapter 4 Applications of Derivatives 



31. 



32. 





33. (a) y' = 16 — x 2 => y' = | +++ | => the curve is rising on (—4, 4), falling on (— oo, —4) and (4, oo) 

-4 4 

=> a local maximum at x = 4 and a local minimum at x = —4; y" = — 2x => y" 



-++| — 




the curve 



is concave up on (— oo, 0), concave down on (0, oo) =4> a point of inflection at x = 



(b) 




34. (a) y' = x 2 - x - 6 = (x - 3)(x + 2) => y' = +++ | | +++ =>• the curve is rising on (-00, -2) and (3, 00), 

-2 3 

falling on (—2, 3) =^ local maximum at x = —2 and a local minimum at x = 3; y" = 2x — 1 

y" — ■ I +++ => concave up on (|, 00) , concave down on (—00, |) =>■ a point of inflection at x = | 



1/2 



(b) 



x=-2 



x=1/2 



x = 3 



35. (a) y' = 6x(x + l)(x - 2) = 6x 3 - 6x 2 - 12x => y' 



the graph is rising on (—1, 0) 



— I +++I I+++ 

-10 2 

and (2, 00), falling on (—00, — 1) and (0, 2) => a local maximum at x = 0, local minima at x = — 1 and 



x = 2; y" = 18x 2 - 12x - 12 = 6 (3x 2 - 2x - 2) = 6 (x - ±=fi\ (x - l - ± fi) 



+++ =>- the curve is concave up on I —00, 



1— </7 1+V7 



i-j/Z^ a„H (1±J1 



and 



/7 A 

-,00 J 



concave down 



l-y/7 l + y/7 



3^- ) =>• points of inflection at x = — y— 



(b) 




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36. (a) y' = x 2 (6 - 4x) = 6x 2 - 4x 3 ^ y' = +++ | +++ | 

3/2 

12x - 12x 2 = 12x(l - x) =>■ y" 



a local maximum at x = | ; y" 



Chapter 4 Practice Exercises 279 
the curve is rising on (— oo, |), falling on (|, oo) 

- =4> concave up on 



I +++ I 
1 



(0, 1), concave down on (— oo, 0) and (1, oo) =>- points of inflection at x = and x = 1 



(b) 



x = 



37. (a) y' = x 4 - 2x 2 = x 2 (x 2 - 2) =4> y' = +- 



-y/2 



V~2 



+++ =>• the curve is rising on ( — oo, — v 2 ) and 



2, oo 1 , falling on ( — y 2, y 2 J =$■ a local maximum at x = — y 2 and a local minimum at x = y 2 ; 

y" = 4x 3 - 4x = 4x(x - l)(x + 1) => y" = | +++ | | +++ => concave up on (-1, 0) and (1, oo), 

-1 1 

concave down on (— oo, —1) and (0, 1) =>• points of inflection at x = and x = ± 1 



(b) 




38. (a) y' = 4x 2 - x 4 = x 2 (4 - x 2 ) => y' 

-c 
4x(2-x 2 ) => y" 



+- 



the curve is rising on (—2, 0) and (0, 2), 



.+++1 — 
-2 2 

falling on (— oo, —2) and (2, oo) => a local maximum at x = 2, a local minimum at x = —2; y" = 8x — 4x 3 



concave up on ( — oo, — y 2 j and (0, y 2 J , concave 
-y/2 ° \[2 

down on I — y 2, J and ( y 2, oo j =^> points of inflection at x = and x = ± y 2 



-+ I — 
V~2 



(b) 



x = 2 




x = -2 



39. The values of the first derivative indicate that the curve is rising on (0, oo) and falling on (— oo, 0). The slope 

of the curve approaches —oo as x — * CT, and approaches oo as x — » + and x — » 1. The curve should therefore 
have a cusp and local minimum at x = 0, and a vertical tangent at x = 1. 



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280 Chapter 4 Applications of Derivatives 




2/3 , „>1/3 
y = x + (x-1 ) 




40. The values of the first derivative indicate that the curve is rising on (0, I) and (1, oo), and falling on (-co, 0) 
and (i, l) ■ The derivative changes from positive to negative at x = i indicating a local maximum there. The 
slope of the curve approaches —oo as x — » 0~ and x — * 1~, and approaches oo as x — » + and as x — > 1 + , 
indicating cusps and local minima at both x = and x = 1. 




2/3 , . x 2/3 
y = x + (x - 1 ) 




41. The values of the first derivative indicate that the curve is always rising. The slope of the curve approaches oo 
as x — > and as x — > 1, indicating vertical tangents at both x = and x = 1, 

y y' 



. 1 -2/3 1 / H v-2/3 



3 








2 








1/3 , , ,1/3 
y = x + (x - 1 ) i 








-3 -2 -1 


^ 1 


2 


3 


-1 








^^^2 








■ """"* -3 










42. The graph of the first derivative indicates that the curve is rising on ( 0, — z ^f — J and ( — ^- — , oo J , falling 
on (—oo, 0) and ( — z ^- — , — ^- — I =4> a local maximum at x = — z ^- — , a local minimum at 
x = — J j^- — . The derivative approaches — oo as x — > 0~ and x — > 1, and approaches oo as x — ► + , 



indicating a cusp and local minimum at x = and a vertical tangent at x = 1 . 



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Chapter 4 Practice Exercises 

y' 



281 



2/3 , H ,1/3 

2 y = x - (x-1) 



-3 -2 



1 2 3 




43. y = |±1 = 1 



4 
x-3 



t 
t+1 _i 1 



44. y = ^_ = 2 - '" 



x + 5 



x + 5 



J\ 


y = 2 






t 


2X 
y_ X + 5 



45. y 



x^ + 1 



46. y 



x^ - X + 1 



X- 1 





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282 Chapter 4 Applications of Derivatives 



47. y 



x J + 2 _ xf , 1 
2x — 2 ' x 



48. y 



x 4 - 1 _ „2 1 




x 3 + 2 _ x* 1 
2x 2 x 




49. y = pEJ = 1 - ^ 



*=-vi 


1 4 

3 

V 2 


t 


x- 2 -3 

j: = ^3 
v=l 










-4 -3 1 


-1 
-1 

-2 

-3 


1 


/2 3 4 




x =-2 x = 2 



51. lim 

x^ 1 



x 2 + 3x - 4 



lim 

x-»l 



2x + 3 
1 



52. lim 

x^ 1 



x a -l 
x b -l 



lim 

x^ 1 



bx b -i b 



53. lim tanx = tan. = 

X — » 7T X 7T 



54. lim 



lim 



x > n x + sinx x > q 1+ cos x 1 + 1 2 



55. lim M% = lim 2 ™ x -, c ,°" = lim -^&, 



lim 



2cos(2x) 



x tiro tan(x 2 ) x 1^ 2xsec2 ( x2 ) x^O 2xsec2 ( x2 ) x^O 2x(2sec 2 (x 2 )tan(x 2 )-2x) + 2sec 2 (x 2 ) + 2-1 



56. lim ^"4 = lim "^"^ = s 

x y n sin(nx) x > n ncos(nx) n 



s(3x) 



x -> tt/2 - 



x -» tt/2 - 



lim 



-3sin(3x) _ 3 



57. lim sec(7x)cos(3x) = lim . — mi. .. . 

..- cos(7x) x— >7r/2~ _7sln ( 7x ) 7 



58. lim ^/x sec x = lim .->— = - = ft 



x^0+ 



x^0 + 



V^ _ 

COS X 1 



59. lim (esc x — cot x) = lim 



1 — COS X 



x^O 



x^O 



lim 

x^ 



sin x 



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Chapter 4 Practice Exercises 283 



60. lim (4 - 4) = lim f 1 ^^ = lim (1 - x 2 ) • 4 = lim (1 - x 2 ) = lim 4 = 1 • oo = oo 

x ^0 Vx4 xV x^O V " 4 / x^O V ; x4 x^O V ; x^O x4 



61. lim (Vx 2 + x+l- Vx 2 -x) = lim (Vx 2 + x + 1 - ^x 2 - xV ^ x2 + x + i+v^ 



lim 



2x+ 1 



X — > OO \A 2 + x + 1 + \/x 2 - x 

Notice that x = y x 2 for x > so this is equivalent to 

2 + i 



lim 

X — > OO 



lim 



v^^+v^ x ^°° v /l + ' + i + v /rr ^ ^^ 



62. lim 

x — » OO 



(_j! xM _ ,• x 3 (x 2 + 1) - xV - 1) _ 

l,x 2 -l ^ + l)~x^OO (x 2 -l)(x 2 + l) - x 



lim 



2x 3 



OO x * 



lim |4 

-> oo ixi 



lim 

-» oo 



12\2 



12 



lim 

x -» oo 24x 



lim ± = 

x ^ OO 2x 



63. (a) Maximize f(x) = y/x- ^36 - x = x 1 / 2 - (36 - x) 1 / 2 where < x < 36 



f'(x) 



-V 2 -U36-xr 1/2 (-l) 



y/36~- 



derivative fails to exist at and 36; f(0) = — 6, 



2^ 2 W " "^ V ^ 2 v /x"\/36-x 

and f(36) = 6 => the numbers are and 36 
(b) Maximize g(x) = ^fx + x/36 - x = x 1 / 2 + (36 - x) 1 / 2 where < x < 36 

critical points at 0, 18 and 36; g(0) = 6, 



> g '(x)=Ix-V 2 + I(36-x)^(-l)=|^ 



2 A '2 

g(18) = 2VT8 = 6v^2 and g(36) = 6 => the numbers are 18 and 18 



64. (a) Maximize f(x) = ^(20 - x) = 20X 1 / 2 - x 3 / 2 where < x < 20 => f'(x) = lOx" 1 / 2 - fx 1 / 2 

_ 20 - 3x _ ^ x = and x = f are critical points; f(0) = f(20) = and f (f ) = Jf (20 - f ) 



2\/x 
3\/3 



40 



the numbers are y and , . 



(b) Maximize g(x) = x + a/20 - x = x + (20 - x) 1 / 2 where < x < 20 =^ g'(x) = ^LzJiz ' 



v/20" 



79 



x = y . The critical points are x = ™ and x = 20. Since g (™) = y and g(20) = 20, 



the numbers must be ™ and J . 



65. A(x) = | (2x) (27 - x 2 ) for < x < \/ri 

=> A'(x) = 3(3 + x)(3 - x) and A''(x) = -6x. 
The critical points are —3 and 3, but —3 is not in the 

domain. Since A"(3) = -18 < and A (y/zj) = 0, 

the maximum occurs at x = 3 => the largest area is 
A(3) = 54 sq units. 




^-1. The 

x z 
v 2 , 128 



66. The volume is V = x 2 h = 32 =>• h 
surface area is S(x) = x 2 + 4x (|f ) = 

where x > => S'(x) 
=4> the critical points are and 4, but is not in the 
domain. Now S"(4) = 2 + ^ > => atx = 4 there 
is a minimum. The dimensions 4 ft by 4 ft by 2 ft 
minimize the surface area. 



2(x - 4) (x 2 + 4x + 16) 
x2 



VIA 



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284 Chapter 4 Applications of Derivatives 



67. From the diagram we have (|) + r 2 = ( y/3 I 
=>• r 2 = 12 ^ h . The volume of the cylinder is 



V = 7rr 2 h = n i 12 ^) h = f (12h - h 3 ) , where 

< h < 2 v^ ■ Then V'(h) = ^ (2 + h)(2 - h) 
=4> the critical points are —2 and 2, but —2 is not in 
the domain. At h = 2 there is a maximum since 
V' (2) = — 3ir < 0. The dimensions of the largest 
cylinder are radius = y/7. and height = 2. 

68. From the diagram we have x = radius and 

y = height = 12 - 2x and V(x) = ± ttx 2 (12 - 2x), where 
< x < 6 =$> V'(x) = 2ttx(4 - x) and V'(4) = -8tt. The 
critical points are and 4; V(0) = V(6) = =4> x = 4 
gives the maximum. Thus the values of r = 4 and 
h = 4 yield the largest volume for the smaller cone. 




Ah 



69. The profit P = 2px + py = 2px + p ( 40 5 _ 1 ® X ) , where p is the profit on grade B tires and < x < 4. Thus 
P'(x) = (5 y. 2 (x 2 - lOx + 20) =>■ the critical points are [5 - V^J . 5, and [5 + y/Tj , but only n - y/S 
the domain. Now P'(x) > for < x < (5 - y/Sj and P'(x) < for (5 - y/yj <x<4=>atx=(5- 
is a local maximum. Also P(0) = 8p, P (5 - y/5j = 4p ( 5 - ^5) w 1 lp, and P(4) = 8p => at x = ( 5 - 

is an absolute maximum. The maximum occurs when x = ( 5 — y/ 5 ) and y = 2 ( 5 — y/ 5 ) , the units are 
hundreds of tires, i.e., x w 276 tires and y « 553 tires. 

70. (a) The distance between the particles is |f(t)| where f(t) = —cos t + cos(t + |) . Then, f'(t) = sin t — sin(t 

Solving f'(t) = graphically, we obtain t « 1.178, t w 4.320, and so on. 

2 



(1.1780972.0 



y/5\ there 
5 ) there 



I)- 




Alternatively, f'(t) = may be solved analytically as follows, f (t) = sin (t + |) - | - sin (t + |) + | 

= ["sin (t + |) cos | -cos(t+ |)sinfl - [sin(t+ |)cosf + cos(t+ |) sin |1 = -2sin|cos(t+ |) 

so the critical points occur when cos(t + f ) = 0, or t = |+ lor. At each of these values, f(t) = ± cos ^ 
w ± 0.765 units, so the maximum distance between the particles is 0.765 units. 



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Chapter 4 Practice Exercises 285 



(b) Solving cos t = cos (t + |) graphically, we obtain t ss 2.749, t w 5.890, and so on. 




Alternatively, this problem can be solved analytically as follows. 

cos t = COS ft + t) 



COS 



[(t+f) 



COS t 



n \ e,;^, TT 



cos(t+ IJcos I + sin(t + l)sin£ = cos(t+ £)cos| - sin(t+ £)sin 
2sin(t + |) sin | = 
sin(t+|) =0 



t 



The particles collide when t 



2.749. (plus multiples of it if they keep going.) 



71. The dimensions will be x in. by 10 - 2x in. by 16 - 2x in., so V(x) = x(10 - 2x)(16 - 2x) = 4x 3 - 52x 2 + 160x for 
< x < 5. Then V'(x) = 12x 2 - 104x + 160 = 4(x - 2)(3x - 20) , so the critical point in the correct domain is x = 2. 
This critical point corresponds to the maximum possible volume because V'(x) > for < x < 2 and V'(x) < for 
2 < x < 5. The box of largest volume has a height of 2 in. and a base measuring 6 in. by 12 in., and its volume is 144 in. 3 
Graphical support: 




72. The length of the ladder is di + d2 = 8 sec 9 + 6 esc 9. We 
wish to maximize 1(9) = 8 sec 9 + 6 esc 9 => I' (9) 
= 8 sec 9 tan 9 - 6 esc 9 cot 9. Then I'(0) = 

2 ^ 



sin 3 61-6 cos 3 9 = 



tan i 



di =4 



36 and d 2 = 3 v / 36 J A + \/36 



the length of the ladder is about 



(^4 + V36) \]a+ 3 v / 36 = (4 + V36) w 19.7 ft 




73. g(x) = 3x-x 3 + 4 => g(2) 
Value Theorem. Then g'(x) = 3 
so forth to x 5 = 2.195823345. 



2>0andg(3) = -14 < 

2 • 3x " 



3x z 



> g(x) 



\ui 



3-3xS 



in the interval [2, 3] by the Intermediate 

; x = 2 => X! = 2.22 => x 2 = 2.196215, and 



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286 Chapter 4 Applications of Derivatives 

74. g(x) = x 4 - x 3 - 75 =>■ g(3) = -21 < and g(4) = 1 17 > =>• g(x) = in the interval [3, 4] by the Intermediate 



3 T v 2 . v _ „ *n ~ *n ~ 75 



Value Theorem. Then g'(x) = 4x 3 — 3x => x n+1 = x n 
=>■ x 2 = 3.229050, and so forth to x 5 = 3.22857729. 



; x = 3 => xi = 3.259259 



75. /(x 3 + 5x - 7) dx = £ + ^ - 7x + C 



76. / (8t 3 - ^ + t ) di = f - $ + '- + C = 21 ' - i- + ^ + C 



77. J(3 v / t+^)dt=/(3t 1 /2 +4 t- 2 )dt 



3tV2 | 4t -l 
(I) 



, C = 2t 3 / 2 - 4 + C 



78 - lfe-?) dt = /(K 1/2 -3t- 4 ) 



dt =H^ -^ + c = ^+? + c 



79. Let u = r + 5 => du = dr 
J RV = J ^ = J u 



du= V + C = -11^+ C 



i +c 

(r + 5) + C 



80. Let u = r - V2 =>■ du = dr 



/^^=6/^-L=6/^=6/u- 3 du = 6(^)+C = -3u- 2 + C 



-V^ 



51. Let u = 6» 2 + 1 =4> du = 2(9 d6» => 1 du = 9 d6 



J 36^6? + 1 dfl=/ A /5(|du) = |/ u 1 / 2 du = § ( s£ ) + C = u 3 /' 2 + C = (9 2 + if 2 + C 



82. Let u = 7 + <9 2 => du = 26» d6» => | du = (9 d6> 



/7fe d0 = /7ua du ) = ^^ 1/2du =Hf +C = ul/2 + C=y7T ^ +C 



83. Let u = 1 + x 4 => du = 4x 3 dx =^ | du = x 3 dx 



/ x 3 (1 + x 4 r 1/4 dx = / U-V4 ( i du) = I J u-V4 du = I f^\ + C = \ u 3 / 4 + C = \ (1 + x 4 ) 3/4 + C 



84. Let u = 2 - x =>• du = - dx => - du = dx 
J (2 - x) 3 / 5 dx = J u 3 / 5 (- du) = - j u 3 / 5 du 



• C = - | u 8 / 5 + C = - | (2 - x) 8 / 5 + C 



85. Let u = ^ =>- du = i ds =>• 10 du = ds 



/see 2 ^ ds = J (sec 2 u) (10 du) = 10 J sec 2 u du = 10 tan u + C = 10 tan ^ + C 

86. Let u = 7rs =>- du = n ds =>• - du = ds 

I esc 2 7ts ds = J (esc 2 u) (i du) = i J esc 2 u du = — i cot u + C = — ^ cot 7rs + C 

87. Letu= v^f? =^ du= Jl&d => 4- du = d <? 

Jcsc v 7 ^ cot 1/20 d6» = J (esc u cot u) (4- du] = 4- (-esc u) + C = - 4- esc \fld + C 



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Chapter 4 Additional and Advanced Exercises 287 



Let u 



du = \ d9 => 3 du = dfl 



J sec | tan f d0 = / (sec u tan u)(3 du) = 3 sec u + C = 3 sec f + C 



Let u = | =4> du = j dx => 4 du = dx 



J sin 2 | dx = J (sin 2 u) (4 du) = J 4 ( '~ c 2 os2u ) du = 2/ (1 - cos 2u) du = 2 (u - ^) + C 
= 2u - sin 2u + C = 2 (|) - sin 2 (|) + C = § - sin | + C 



90. Let u = | =4> du = i dx => 2 du = dx 



J cos 2 § dx = J (cos 2 u) (2 du) = J*2 ( 1+c 2 os2u ) du = / (1 + cos 2u) du = u + ^ + C 



f + \ sin x + C 



91. y = J ^tl dx= J(l + x- 2 ) dx = x - x- 1 + C = x- ; + C; y = -1 when x = 1 =^> l-± + C = -l 

=► C=-l =► y = x-i-l 

92. y = J (x + i) 2 dx = J (x 2 + 2 + 4j) dx = J (x 2 + 2 + x~ 2 ) dx = f + 2x - x" 1 + C = f + 2x - 1 + C; 



y = 1 when x=l => f+2-{+C=l => C 



y - 3 -f zx x 3 



93. 



di 



/ (l5v/t + X)dt= J (151 1 / 2 + 31- 1 / 2 ) dt = 10t 3 / 2 + 6t : / 2 + C; f = 8 when t = 1 

=> 10(1) 3 / 2 + 6(1) J / 2 + C = 8 => C = -8. Thus g = 10t 3 / 2 + 61 1 / 2 - 8 => r = J (lOt 3 / 2 + 6I 1 / 2 - 8) dt 
= 4t 5 /2 + 4t 3/2 _ 8t + C; r = when t = 1 => 4(1) 5 / 2 + 4(1) 3 / 2 - 8(1) + d = =>• Ci = 0. Therefore, 
r = 4t 5 / 2 + 4t 3 / 2 - 8t 



94. |f = J -cos t dt = -sin t + C; r" = when t = =4> -sin + C = =4> C = 0. Thus, |f = -sin t 
=> % = /-sintdt= cost + Ci;r' = whent = =>■ 1 + Ci = => Ci = -1. Then | = cos t - 1 
=>• r = J (cos t - 1) dt = sin t - t + C 2 ; r = -1 when t = => - + C 2 = -1 =>■ C 2 = -1. Therefore, 
r = sin t — t — 1 

CHAPTER 4 ADDITIONAL AND ADVANCED EXERCISES 

1 . If M and m are the maximum and minimum values, respectively, then m < f(x) < M for all x e I. If m = M 
then f is constant on I. 



2. No, the function f(x) 



3x + 6, -2 < x < 
9 - x 2 , < x < 2 



has an absolute minimum value of at x = —2 and an absolute 



maximum value of 9 at x = 0, but it is discontinuous at x = 0. 

3. On an open interval the extreme values of a continuous function (if any) must occur at an interior critical 
point. On a half-open interval the extreme values of a continuous function may be at a critical point or at the 
closed endpoint. Extreme values occur only where f ' = 0, f ' does not exist, or at the endpoints of the interval. 
Thus the extreme points will not be at the ends of an open interval. 



4. The pattern ? = +++ 



minimum at x = 3. 



-++ indicates a local maximum at x = 1 and a local 



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288 Chapter 4 Applications of Derivatives 

5. (a) If y' = 6(x+ l)(x - 2) 2 , theny' < Oforx < -1 and y' > Oforx > -1. The sign pattern is 

f = | +++ | +++ => fhas a local minimum at x = -1. Also y" = 6(x - 2) 2 + 12(x + l)(x - 2) 

-1 2 

= 6(x — 2)(3x) =>■ y'' > for x < or x > 2, while y" < for < x < 2. Therefore fhas points of inflection 
at x = and x = 2. There is no local maximum, 
(b) If y' = 6x(x+ l)(x - 2), then y' < for x < -1 and < x < 2; y' > for -1 < x < and x > 2. The sign 
sign pattern is y' = — +++ | — | +++ . Therefore fhas a local maximum at x = and 



-1 



local minima at x = -1 and x = 2. Also, y" = 18 |x - (^^Hl I* - ( ii 3^)l ' so f < ° 
1—fl < x < H^lL and y" > for all other x => f has points of inflection at x = ^fl 



for 



6. The Mean Value Theorem indicates that ^ijf* = f'(c) < 2 for some c in (0, 6). Then f(6) - f(0) < 12 
indicates the most that f can increase is 12. 

7. If f is continuous on [a, c) and f '(x) < on [a, c), then by the Mean Value Theorem for all x e [a, c) we have 
f "^lf° < => f(c) - f(x) < => f(x) > f(c). Also if f is continuous on (c, b] and f'(x) > on (c, b], then for 
all x G (c, b] we have f(x) : f(c) > =4> f(x) - f(c) > => f(x) > f(c). Therefore f(x) > f(c) for all x G [a, b]. 



(a) For all x, -(x + l) 2 < < (x - l) 2 => - (1 + x 2 ) < 2x < (1 + x 2 ) =» - \ < jf^ < \ . 

jf^ \<\, from part (a) 



(b) There exists c G (a, b) such that -^ = f( ^:f° 



f(b)-f(a) 
b-a 



|f(b) - f(a)| < i |b - a| 



9. No. Corollary 1 requires that f '(x) = for all x in some interval I, not f '(x) = at a single point in I. 

10. (a) h(x) = f(x)g(x) =>■ h'(x) = f'(x)g(x) + f(x)g'(x) which changes signs at x = a since f'(x), g'(x) > when 

x < a, f'(x), g'(x) < when x > a and f(x), g(x) > for all x. Therefore h(x) does have a local maximum 
at x = a. 
(b) No, let f(x) = g(x) = x 3 which have points of inflection at x = 0, but h(x) = x 6 has no point of inflection 
(it has a local minimum at x = 0). 



11. From(ii), f(-l) 



b-c + 2 







x+1 



lim__f(x) = ^ lim__ bx2 + cx + 2 



X — ► ± oo % x^±oo 

lim 

x^ ±oc x + c + 



a = 1; from (iii), either 1 = lim f(x) or 1 = lim f(x). In either case, 

lim r— ; — *r 
x^±oo bx + c + : 



and if c = 0, then lim 

X -> ± OO bx + 7 



1 =$■ b = and c = 1. For if b = 1, then 



lim 

X — > ± 00 



± oo. Thus a = 1, b = 0, and c = 1. 



12. | = 3x 2 + 2kx + 3 = => x 
k= ±3. 



-2k± y/4k 2 -36 
6 



x has only one value when 4k — 36 = => k = 9 or 



13. The area of the AABC is A(x) = \ 
where < x < 1. Thus A'(x) 



(2)VT 



yr 



- X 2 = ( 1 — X 

and ± 1 are 



2N1/2 



critical points. Also A ( ± 1) = so A(0) = 1 is the 
maximum. When x = the AABC is isosceles since 
AC = BC= \/l. 



(x.Vl-x 2 )c 




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Chapter 4 Additional and Advanced Exercises 289 



14. lim 



f'(c + h)-f'(c) 







f"(c) =>• for e = \ |f"(c)| > there exists a 6 > such that < |h| < 6 



f (c+h) - f'(c) 



- f"(c) < i |f"(c)| . Then f'(c) = 



£ |f"(c)| < 



f'(c + h) 



-f"(c)<i|f"(c)| 



=> f"(c) - 1 |f"(c)| < ^^ < f"(c) + i |f"(c)| . If f"(c) < 0, then |f"(c)| = -f"(c) 



| f'( c ) < Li£±h) < i f '( C ) < 0; likewise if f "(c) > 0, then < \ f"(c) < ^£ + W < | f"( c ) 



h ^2 

f '(c + h) < 0. Therefore, f(c) is a local 



(a) If f "(c) < 0, then -6 < h < => f '(c + h) > and < h < 6 
maximum. 

(b) If f "(c) > 0, then -6 < h < => f'(c + h)< and < h < 6 =>■ f'(c + h) > 0. Therefore, f(c) is a local 
minimum. 



15. The time it would take the water to hit the ground from height y is < / — , where g is the acceleration of 



gravity. The product of time and exit velocity (rate) yields the distance the water travels: 



D(y) 



v/64(h - y) = 8 J\ (hy - y 2 ) ' , < y < h => D'(y) = -4 J\ (hy - y 2 ) ' (h - 2y) => 0, | and h 



,21-1/2, 



are critical points. Now D(0) = 0, D i 



8h 



and D(h) = =>• the best place to drill the hole is at y 



16. From the figure in the text, tan (/3 + 6) = ^±»; tan (/3 + 9) = ^^aZ*e \ and tan 9 = \ . These equations 

tan/3+j htan,fl + a 



give^ 



1 



tan/3 



h — a tan Q 



Solving for tan /3 gives tan f3 



bh 

h 2 + a(b + a) 



or 



(h 2 — a(b + a)) tan j3 = bh. Differentiating both sides with respect to h gives 

2h tan P + (h 2 + a(b + a)) sec 2 (3 % = b. Then f = =► 2h tan /3 = b => 2h ( h2 + a b ( h b + a) ) 



2bh 2 = bh 2 + ab(b + a) 



h 2 



i(b + a) => h = v /a(a~+bj . 



17. The surface area of the cylinder is S = 27rr 2 + 27rrh. From 



the diagram we have g = ^pp => h = RH R rIi and 
S(r) = 2?rr(r + h) = 2?rr (r + H - r g) 
= 2?r (l - |) r 2 + 27rHr, where < r < R. 
Case 1: H < R =^> S(r) is a quadratic equation containing 

the origin and concave upward =>■ S(r) is maximum at 

r = R. 
Case 2: H = R => S(r) is a linear equation containing the 

origin with a positive slope => S(r) is maximum at 

r = R. 
Case 3: H > R => S(r) is a quadratic equation containing the origin and concave downward. Then 




dS 
di 

we let r* 



4tt (1 - |) r + 2ttH and f = => 4tt (l - | ) r + 2ttH = => r : 



RH 



2(H-R) 



For simplification 



RH 



2(H - R) 



(a) If R < H < 2R, then > H - 2R => H > 2(H - R) => 2 (h^r) > R which is im P ossible 



(b) IfH = 2R, thenr* 



2R 2 
2R 



R 



S(r) is maximum at r = R. 



(c) If H > 2R, then 2R + H < 2H => H < 2(H - R) 



H 



2(H - R) 



< 1 => 



RH 



2(H - R) 



<R 



r* < R. Therefore, 



RH 



S(r) is a maximum at r = r* = jt^ 



R) 



Conclusion : If H € (0, R] or H = 2R, then the maximum surface area is at r = R. If H e (R, 2R), then r > R 
which is not possible. If H 6 (2R, oo), then the maximum is at r = r* = 2(I RH R . 



18. f(x) = mx - 1 + i => f'(x) = m - \ and f'(x) = \ > when x > 0. Then f'(x) = => x = -J- yields a 

XXX V IT1 

l. If f ( -t= ) > 0, then A^/m — 1 + ^/m = 2^/m — 1 > => m>i. Thus the smallest acceptable value 



minimum. 



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290 Chapter 4 Applications of Derivatives 



for m is ^ . 



2sin(5x) 



19. (a) lim 

W x^O 3x 



lim 



2sin(5x) 



X^O §( 5x ) X^O 3 ( 5x > 



lim &*°j& = m.i = i& 



(b) lim sin(5x)cot(3x) = lim sin(5x ' c °f x) 
W x^O K ' V ' x^O Sln < 3x ) 



lim 



]• — 3sin(5x)sin(3x) + 5cos(5x)cos(3x) 5 

x __j q 3cos(3x) — 3 



(c) lim x esc V2x = lim — r 5 

x -> x -» sin 2 ^ 



lim 



lim 



2x x ^0 asm^a^^ x _, sin ^ 2 /2x") x^O cos(2v^x) 



2_ 



= lim — t — =s— = 

X^O cos (2^/2x1 -2 

(d) lim (sec x — tan x) 

X — > 7T/2 



X — > 

1 — COS x 



lim ±=^ 

-> tt/2 cos x 



lim =^ 

x^tt/2 - smx 



(e) lim ^f^ = lim 

v ' x j. n x — tan x x r\ 1 — sec^x 



lim 

x^O 



lim 55^ 

x -> tan x 



lim 9 - Slnx 2 

x j, n 2 tan x sec^x 



lim 



,, 2 s i n x 
cos^x 



= lim £2^ 

x^O ~ 2 

(f) lim ^1 = 
x -> xsln x 



2x cos(x 2 



(g) x lim ^Jp 



2x 



x^O 

,. (x-2)(x 2 + 2x + 4) _ 

x lml 2 (x-2)(x + 2) -, 



-(2x 2 )sin(x 2 ) + 2cos(x 2 ) _ 2 
— xsin x+2cos x 2 

i* rn sec 3 x + tan 2 x sec x 1 + 1 



lim ^xcus > x-; _ jj m -^x-^sunx-^i-zcos^A-; _ 2 _ ^ 

x > a xcos x+sin x x > n —xsin x+2cos x 2 

ii rn sec x tan x 

~ x^O 



lim x+2x + 4 — 4 + 4 + 4 

-> 2 



x+2 



20. (a) lim 



yx + 5 

V^+5 



lim 



(b) lim 



2x 
6o x + 7,/x 



CO yA + 5 
2x 



lim 



/1 + a 



X — > CO 1 ■ 



lim 



2 



CO x + 7 \/ x 



lim 

l ^°° 1+7, A 



2 

1+0 



21. (a) The profit function is P(x) = (c — ex)x — (a + bx) = —ex 2 + (c — b)x — a. P'(x) = — 2ex + c — b = 



2e 



. P"(x) = — 2e < Oif e > so that the profit function is maximized at x 



c-b 
2e ' 



(b) The price therefore that corresponds to a production level yeilding a maximum profit is 



c — e 



2e 



dollars. 



(c) The weekly profit at this production level is P(x) = -e(^) + (c - b)(- c_1> ] 

(d) The tax increases cost to the new profit function is F(x) = (c — ex)x — (a + bx + tx) = —ex 
Now F'(x) = -2ex + c - b - t = when x = ^p = ' L ^ 1 - Since F"(x) = -2e < if e > 0, F is maximized 



a=^-a. 

2 -J- <c - b - t)x - a. 



when x 



2e 



units per week. Thus the price per unit is p = c — e( c 2 ^ ' ) 



c + b + t 



dollars. Thus, such a tax 



increases the cost per unit by c+ ^ +t — £ ^ = | dollars if units are priced to maximize profit. 



22. (a) 




The x-intercept occurs when i — 3 = 0=>^=3=^x=i. 
(b) By Newton's method, x n+1 = x n - jS|4-. Here f'(x n ) = -x~ 2 = ^. So x n+1 = x n 
= x n + x n ox n = ZX n ox n = X n ^Z oX n J. 



Xn + 



--3Vn 



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Chapter 4 Additional and Advanced Exercises 291 



f(*o) 
f'(*o) 



X 



o~ a _ qxg - xg - a _ xg(q-l)-a 

i x 3 



q^" 1 



q^" 1 



— | — — Xq ( - — ) + -5=1 I - J so that xi is a weighted average of Xq 



and -^=j with weights mo 



q-i 



and nil = - . 

q q 



In the case where x„ = ^we have x q = a and Xl = ^ (*=i) + -^ (l) = -^ (a_i + l) 



a 

x o 



24. We have that (x - h) z + (y - h) z = r 2 and so 2(x - h) + 2(y - h)g = and 2 + 2^ + 2(y - h)3J = hold 



>dy 



,<fy 



rfi 



,dy 



Thus 2x + 2y2i = 2h + 2hf, by the former. Solving for h, we obtain h 



x + y — 

jjr ■ Substituting this into the second 



equation yields 2 + 2^ + 2y 



ax 2 



*±y£ 



0. Dividing by 2 results in 1 



dy i v d!z 

dx T J dx 2 



*+yg 



25. (a) a(t) = s"(t) = -k (k > 0) => s'(t) = -kt + Cj, where s'(0) 



Ci = 88 => s'(t) 



-kt 



s(t) 

-kt 2 
2 



-kt 2 

2 



C 2 where s(0) = =>■ C 2 = so s(t) 



-kt 2 



., + 88t. Now s(t) = 100 when 
+ 88t = 100. Solving for t we obtain t = 88 ± ^f ~ 200k . At such t we want s'(t) = 0, thus 



' 88 + y^ 2 - 200k " 



_ K / of -h y oo~ — ZUUK \ 



Oor kf 88 -^ 8 f- 200k ) 



So 



so that k = |§g w 38.72 ft/sec 2 . 
(b) The initial condition that s'(0) = 44 ft/sec implies that s'(t) 



0. In either case we obtain ! 



-kt 2 



5 2 



200k = 



-kt + 44 and s(t) = -jp + 44t where k is as above. 



The car is stopped at a time t such that s'(t) = — kt + 44 = => t = y . At this time the car has traveled a distance 



< i lt) = f(f) +44(f) = f 
stopping distance. 



k 



968 (fj§) = 25 feet. Thus halving the initial velocity quarters 



26. h(x) = f 2 (x) + g 2 (x) => h'(x) = 2f(x)f'(x) + 2g(x)g'(x) = 2[f(x)f'(x) + g(x)g'(x)] = 2[f(x)g(x) + g(x)(-f(x))] 
= 2-0 = 0. Thus h(x) = c, a constant. Since h(0) = 5, h(x) = 5 for all x in the domain of h. Thus h(10) = 5. 



27. Yes. The curve y = x satisfies all three conditions since -=| = 1 everywhere, when x = 0, y = 0, and -^ 



everywhere. 



28. y' = 3x 2 + 2 for all x =>■ y = x 3 + 2x + C where -l = l 3 + 2-l + C=>C = -4^y = x 3 + 2x-4. 



29. s"(t) = a = -t 2 =>• v = s'(t) 



C. We seek Vq = s'(0) = C. We know that s(t*) = b for some t* and s is at a 



maximum for this t*. Since s(t) = -£ + Ct + k and s(0) = we have that s(t) = -£ + Ct and also s'(t*) = so that 



t* = (3C) 



1/3 - [-(3C) 1 / 3 ] 4 



So 



C(3C) 



1/3 



ubr 



Thus v = s'(0) 



(4b) J 



b^(3C) 1/3 (C-§) 



12 

(3C) 1/3 (f)=b^3 1 /3 C 4 / 3 =f 



30. (a) s"(t) = t 1 / 2 - r 1 ! 2 =>■ v(t) = s'(t) = ft 3 / 2 - 2I 1 / 2 + k where v(0) 

4 



K ~ 3 

J_t5/2 _ l t 3/2 
15 l 3 l 



v ( t ) = | t 3/2 _ 2t l/2 + 4 



3 l 15 ' 



(b) s(t) = j|t 5 / 2 - ft 3 / 2 + ft + k 2 where s(0) = k 2 = -±. Thus s(t) 
31. The graph of f(x) = ax 2 + bx + c with a > is a parabola opening upwards. Thus f(x) > for all x if f(x) = for at most 



one real value of x. The solutions to f (x) = are, by the quadratic equation 

(2b) 2 - 4ac < => b 2 - ac < 0. 



:yW~' 



Thus we require 



32. (a) Clearly f(x) = (aix +bi) + . . . + (a n x + b n ) > for all x. Expanding we see 



f(x) = (a 2 x 2 + 2 ai b lX + b 2 ) + . . . + (a 2 x 2 + 2a n b n x + b 2 ) 
= (a 2 + a 2 + . . . + a 2 )x 2 + 2(aib a + a 2 b 2 + . . . + a n b n )x + (b 2 + b 
Thus (aibj + a 2 b 2 + . . . + a n b n ) 2 - (a 2 + a 2 + . . . + a 2 )(b 2 + b 2 + 



• • + b 2 ) > 0. 

b 2 ) < by Exercise 31. 



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292 Chapter 4 Applications of Derivatives 



Thus (aibi + a 2 b 2 + . . . + a n b n ) 2 < (af + a| + . . . + ajj)(b? + b 2 + . . . + bjj). 
(b) Referring to Exercise 31: It is clear that f (x) = for some real x O b 2 — 4ac = 0, by quadratic formula. 
Now notice that this implies that 



f(x) = (aix + b : ) 2 + . . . + (a„x + b n ) 2 



= (a 2 + a 2 + . . 
<^> (aibi + a 2 b 2 + 
<^> (aibi + a 2 b 2 + 



2(aibi + a 2 b : 



2°2 



a n b n )x + (b 2 



bl 



b) 







+ a n b n ) 2 - (a 2 + a 2 + . . . + a 2 )(b 2 + b 



ar,b n 



(a 2 + a 2 + ... + a 2 )(b 2 +b 2 +.. 



But now f(x) = •£>• a,x + b; = for all i = 1, 2, . . . , n •«• a;x ■ 



-bi 



+ b 2 ) = 
+ b 2 ) 
= for all i 



1,2, 



n. 



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CHAPTERS INTEGRATION 



5.1 ESTIMATING WITH FINITE SUMS 




(a) Ax = ^ = \ and X; = iAx 



(b) Ax = i^ = \ and x; = iAx 



Since f is increasing on [0, 1], we use left endpoints to obtain 
lower sums and right endpoints to obtain upper sums. 



a lower sum is 



E(i) , -l = l(o , + (l) , ) = i 



(c) Ax = iyS = i and x; = iAx = | =^> an upper sum is $^(5) ' " = " ' ^ 4 ' 

i=l 



i-n 

alowersumisE(t) 2 -| = U° 2+ (3) 2+ (5) 2 +(I) 2 ) 
i=o v ' 

2) ' 2 ~~ 2 \ V2V T± / — 8 



I I 
4 ' 8 



J7_ 

■ VI 



(d) Ax 



i-o _ 1 



I and x; = iAx = \ =4> an upper sum is ^ ( 



l _ l / 1\ 



:d 2 +(!) 2 ^ 



'30^ 
v 16 J 



15 

■ VI 




Since f is increasing on [0, 1], we use left endpoints to obtain 
lower sums and right endpoints to obtain upper sums. 



(a) Ax = Ljp = \ and x; = iAx = \ 



a lower sum is 



(b ) Ax=i^ = i 

(c) Ax = iffl = 1 and x } = iAx = | 

(d) Ax = i=o = i and x . = iAx = i 



Ea)'4=i(o"+(i)')=^ 

i=0 v y 

andx, = iAx = \ ^ a lower sum is £(i) 3 ■ \ = | (o 3 + (i) 3 + Q) 3 + (f) 3 ) 

i=o v 7 

uppersumisX:(i) 3 -i = |(G) 3 +1 3 ) = £ • | = & 

-g(i) 3 -! = i(a) 3 +(D 3 +(i) 3 +i 3 ) 



an upper sum i 



36 __ 9 

256 64 



100 _ 25 
256 64 



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294 Chapter 5 Integration 



3. f(x) 



Since f is decreasing on [0, 1], we use left endpoints to obtain 
upper sums and right endpoints to obtain lower sums. 



(a) Ax = £f± = 2 and xi = 1 + iAx = 1 + 2i =>■ a lower sum is £ j: ' 2 = 2 (| + £ 



i-1 
4 



15 



(b) Ax = ^j 1 = 1 and xi = 1 + iAx = 1 + i =>- a lower sum is^F' 1 = 1 (5 + 5 + 3 + 5 



i=l 



1 i 1 i 1 i 1> _ II 
60 



(c) Ax 



5-1 



2 and x; = 1 + iAx = 1 + 2i => an upper sum is ^^ • 2 = 2(l + 4 



i=0 
3 



(d) Ax = ^-j 1 = 1 and x, = 1 + iAx = 1 + i =>• an upper sum is^F' 1 = 1 ( 1 +5 + 5 + 3) = ff 



i=0 



4. f(x) = 4 - x 2 




Since f is increasing on [—2, 0] and decreasing on [0, 2], we use 
left endpoints on [—2, 0] and right endpoints on [0, 2] to obtain 
lower sums and use right endpoints on [—2. 0] and left endpoints 
on [0, 2] to obtain upper sums. 



(a) Ax = 2 [ 2) = 2 and x ; = -2 + iAx = -2 + 2i =>• a lower sum is 2 • (4 - (-2) 2 ) + 2 • (4 - 2 2 ) = 

1 4 

(b) Ax = ^t^ = 1 and x ; = -2 + iAx = -2 + i => a lower sum is ]£ (4 - (x ; ) 2 ) • 1 + ]T (4 - (x ; ) 2 ) • 1 

i=0 i=3 

= 1((4 - (-2) 2 ) + (4 - (-1) 2 ) + (4 - l 2 ) + (4 - 2 2 )) = 6 

(c) Ax = 2 ~^~ 2) = 2 and x ; = -2 + iAx = -2 + 2i => a upper sum is 2 • (4 - (0) 2 ) + 2 • (4 - 2 ) = 16 



(d) Ax 



2 -(-2) 



1 and x; = -2 + iAx = -2 + i => a upper sum is £ (4 - (x ; ) 2 ) • 1 + ]T (4 - (x ; ) 2 ) • 1 



i=l 



3 

— r 

i-2 



1((4 - (-1) 2 ) + (4 - 2 ) + (4 - 2 ) + (4 - l 2 )) = 14 




Using 2 rectangles =>• Ax 



¥ = }=>iWi)m)) 



l((i) 2 + (!) 2 ) 



10 _ _5_ 
32 16 



1-0 _ 1 

4 4 



Using 4 rectangles => Ax 
=>i(f(l)+f(|) + f(|)+f(|)) 

= 4-(Q) 2 + (f) 2 + (I) 2 + (f) 2 ) = ! 



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Section 5.1 Estimating with Finite Sums 295 




Using 2 rectangles 


^Ax = 


1-0 _ 

2 


1 
2 


= 2-((3) 3 + (D 3 


\ 28 
) 2-64 


7 
32 




Using 4 rectangles 


=> Ax = 


1-0 _ 

4 


1 

4 


=►*(*(*) +f(§) 


+f(l)+f(D) 




1 ( l 3 +3 3 +5 3 +7 3 A 
~ 4 \ 8 3 j 


496 

~~ 4-8 3 ~~ 


124 

8 3 — 


31 

12.x 



l(f(i)+f(!)) 



7. f(x) = i 
y 



f(x) = 4 - x 2 



—i 1 — ►- x 

4 5 




Jsing 2 rectangles =>• Ax 


= 5^1 =2^2 


(f(2) 


+ f(4)) 


= 2(| + 1) = | 








Jsing 4 rectangles => Ax 


= ^ = 1 






^l(f(l)+f(l)+f(I) 


+f(D) 






= 1( 2 + 2 + 2 + 2-) = 


1488 _ 496 _ 
■5-7-9 5-7-9 


496 
315 




Jsing 2 rectangles => Ax 


- 2 - { - 2) -2^2(f( 


-l) + f(l)) 


= 2(3 + 3) = 12 








Jsing 4 rectangles => Ax 


_ 2- (-2) _ 1 
4 - 1 - 






=M(f(-|)+f(-|)+f 


Q)+f(D) 






= l((4-(-l) 2 ) + ( 4 


-H) 2 ) + ( 4 


~{\ 


)V(4- 


= 16 -(| -2 + 1-2) = 


16 -f = 11 







! )) 



9. (a) D « (0)(1) + (12)(1) + (22)(1) + (10)(1) + (5)(1) + (13)(1) + (H)(1) + (6)(1) + (2)(1) + (6)(1) = 87 inches 
(b) D « (12)(1) + (22)(1) + (10)(1) + (5)(1) + (B)(1) + (H)(1) + (6)(1) + (2)(1) + (6)(1) + (0)(1) = 87 inches 

10. (a) D « (1)(300) + (1.2)(300) + (1.7)(300) + (2.0)(300) + (1.8)(300) + (1.6)(300) + (1.4)(300) + (1.2)(300) 

+ (1.0)(300) + (1.8)(300) + (1.5)(300) + (1.2)(300) = 5220 meters (NOTE: 5 minutes = 300 seconds) 
(b) D « (1.2)(300) + (1.7)(300) + (2.0)(300) + (1.8)(300) + (1.6)(300) + (1.4)(300) + (1.2)(300) + (1.0)(300) 
+ (1.8)(300) + (1.5)(300) + (1.2)(300) + (0)(300) = 4920 meters (NOTE: 5 minutes = 300 seconds) 

11. (a) D « (0)(10) + (44)(10) + (15)(10) + (35)(10) + (30)(10) + (44)(10) + (35)(10) + (15)(10) + (22)(10) 

+ (35)(10) + (44)(10) + (30)(10) = 3490 feet « 0.66 miles 
(b) D « (44)(10) + (15)(10) + (35)(10) + (30)(10) + (44)(10) + (35)(10) + (15)(10) + (22)(10) + (35)(10) 
+ (44)(10) + (30)(10) + (35)(10) = 3840 feet « 0.73 miles 

12. (a) The distance traveled will be the area under the curve. We will use the approximate velocities at the 

midpoints of each time interval to approximate this area using rectangles. Thus, 

D « (20)(0.001) + (50)(0.001) + (72)(0.001) + (90)(0.001) + (102)(0.001) + (112)(0.001) + (120)(0.001) 
+ (128)(0.001) + (134)(0.001) + (139)(0.001) « 0.967 miles 
(b) Roughly, after 0.0063 hours, the car would have gone 0.484 miles, where 0.0060 hours = 22.7 sec. At 22.7 
sec, the velocity was approximately 120 mi/hr. 



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296 Chapter 5 Integration 

13. (a) Because the acceleration is decreasing, an upper estimate is obtained using left end-points in summing 
acceleration • At. Thus, At = 1 and speed w [32.00 + 19.41 + 11.77 + 7.14 + 4.33](1) = 74.65 ft/sec 

(b) Using right end-points we obtain a lower estimate: speed rs [19.41 + 11.77 + 7.14 + 4.33 + 2.63](1) 

= 45.28 ft/sec 

(c) Upper estimates for the speed at each second are: 



t 





1 


2 


3 


4 


5 


V 





32.00 


51.41 


63.18 


70.32 


74.65 



Thus, the distance fallen when t = 3 seconds is s w [32.00 + 51.41 + 63.18](1) = 146.59 ft. 



14. (a) The speed is a decreasing function of time 
attained. Also 



right end-points give an lower estimate for the height (distance) 



t 





1 


2 


3 


4 


5 


V 


400 


368 


336 


304 


272 


240 



gives the time-velocity table by subtracting the constant g = 32 from the speed at each time increment 
At = 1 sec. Thus, the speed w 240 ft/sec after 5 seconds, 
(b) A lower estimate for height attained is h w [368 + 336 + 304 + 272 + 240](1) = 1520 ft. 



15. Partition [0,2] into the four subintervals [0,0.5], [0.5, 1], [1, 1.5], and [1.5,2]. The midpoints of these 
subintervals are mi = 0.25, iri2 = 0.75, ni3 = 1.25, and mi = 1.75. The heights of the four approximating 
rectangles are f( mi ) = (0.25) 3 = ^ f(m 2 ) = (0.75) 3 = g, f(m 3 ) = (1.25) 3 = if, and f(m 4 ) = (1.75) 3 - ^ 



Notice that the average value is approximated by \ 



:d'(§; 



;!)'(§; 



" 64 



64 



:i) 3 (i: 



31 

16 



length of [0,2] 



approximate area under 
curve f(x) = x 3 



We use this observation in solving the next several exercises. 



16. Partition [1 , 9] into the four subintervals [1, 3], [3, 5], [5, 7], and [7, 9]. The midpoints of these subintervals are 
mi = 2, ni2 = 4, 1113 = 6, and mi = 8. The heights of the four approximating rectangles are f(mi) = \, 



f(m2) = t, f(ni3) = |, and f(iri4) = =, The width of each rectangle is Ax = 2. Thus, 



Area 



2 a) 



2( )+2Q; 



25 
12 



average value : 



area 
length of [1,9] 



(I) 
8 



25 
96" 



17. Partition [0,2] into the four subintervals [0,0.5], [0.5, 1], [1, 1.5], and [1.5,2]. The midpoints of the subintervals 
are mi = 0.25, m 2 = 0.75, m 3 = 1.25, and m 4 = 1.75. The heights of the four approximating rectangles are 

1 c ;„2 5ir _ 1 



f(mi) 



1 I _* 2 7T 

tj + sin -r 



l,f(m 2 ) 



\ + sin 2 2f 






l,f(m 3 ) 



k} 



v^ 



= j + ^ = 1, and f(m4) — , 
Area«(l + l + l + l)(i) =2 



1+sin2 T = l + (-^)^ 



average value 



length of [0,2] 



1 . The width of each rectangle is Ax 
1 = 1. 



i Thus, 



18. Partition [0, 4] into the four subintervals [0, 1], [1,2, ], [2, 3], and [3,4]. The midpoints of the subintervals 
are mi = | , m 2 = | , m 3 = | , and m 4 = | . The heights of the four approximating rectangles are 

f(mi) = 1 - (cos (^ir)) = l ~ ( cos (I)) 4 = 0-27145 (to 5 decimal places), 

f(m 2 ) = 1 - (cos (^)) = 1 - (cos (f )) 4 = 0.97855, f(m 3 ) = 1 - (cos (^)) = 1 - (cos (f )) 4 

= 0.97855, and f(m 4 ) = 1 - (cos (^)) = 1 - (cos(^)) 4 = 0.27145. The width of each rectangle is 

Ax=l. Thus, Area « (0.27145)(1) + (0.97855)(1) + (0.97855)(1) + (0.27145)(1) = 2.5 => average 
value 



area 
length of [0,4] 



225 _ 5 
4 — 8' 



Copyright (c) 2006 Pearson Education 




Section 5.1 Estimating with Finite Sums 297 

19. Since the leakage is increasing, an upper estimate uses right endpoints and a lower estimate uses left 
endpoints: 

(a) upper estimate = (70)(1) + (97)(1) + (136)(1) + (190)(1) + (265)(1) = 758 gal, 
lower estimate = (50)(1) + (70)(1) + (97)(1) + (136)(1) + (190)(1) = 543 gal. 

(b) upper estimate = (70 + 97 + 136 + 190 + 265 + 369 + 516 + 720) = 2363 gal, 
lower estimate = (50 + 70 + 97 + 136 + 190 + 265 + 369 + 516) = 1693 gal. 

(c) worst case: 2363 + 720t = 25,000 =>■ t«31.4hrs; 
best case: 1693 + 720t = 25,000 =>■ t w 32.4 hrs 

20. Since the pollutant release increases over time, an upper estimate uses right endpoints and a lower estimate 
uses left endpoints: 

(a) upper estimate = (0.2)(30) + (0.25)(30) + (0.27)(30) + (0.34)(30) + (0.45)(30) + (0.52)(30) = 60.9 tons 
lower estimate = (0.05)(30) + (0.2)(30) + (0.25)(30) + (0.27)(30) + (0.34)(30) + (0.45)(30) = 46.8 tons 

(b) Using the lower (best case) estimate: 46.8 + (0.52)(30) + (0.63)(30) + (0.70)(30) + (0.81)(30) = 126.6 tons, 
so near the end of September 125 tons of pollutants will have been released. 

21. (a) The diagonal of the square has length 2, so the side length is y 2. Area = 

(b) Think of the octagon as a collection of 16 right triangles with a hypotenuse of length 1 and an acute angle measuring 

2tt TV 

16 — 8' 

Area = 16fJ) (sin |) (cos |) = 4 sin f = 2 a/2 « 2.828 

(c) Think of the 16-gon as a collection of 32 right triangles with a hypotenuse of length 1 and an acute angle measuring 

2tt _ jr_ 
32 16 - 



Area = 32(i) (sin i) (cos ^] 



sin 



2^2 w 3.061 



(d) Each area is less than the area of the circle, n. As n increases, the area approaches ir. 



22. (a) Each of the isosceles triangles is made up of two right triangles having hypotenuse 1 and an acute angle measuring 
|| = |. The area of each isosceles triangle is At = 2(g) (sin |) (cos ^) = | sin ^. 

(b) The area of the polygon is Ap = nAx = § sin — , so lim § sin — = lim tt ■ -j^f- = n 

1 n n^oo^ n n^oo \T) 

(c) Multiply each area by r 2 . 





A T = ir 2 sin — 

1 2 n 




A P = 2r 2 sin 2l 

r 2 n 




limAp = 7rr 2 

n— >oo 


-26. 


Example CAS commands: 


Maple: 




with( Student[Calculusl] ); 



f := x -> sin(x); 

a:=0; 

b := Pi; 

plot( f(x), x=a..b, title="#23(a) (Section 5.1)" ); 

N:=[100, 200, 1000]; 

for n in N do 

Xlist := [ a+l.*(b-a)/n*i $ i=0..n ]; 

Ylist := map( f, Xlist ); 
end do: 
for n in N do 

Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist)); 



#(b) 



#(c) 



Copyright (c) 2006 Pearson Education 




298 Chapter 5 Integration 

end do; 

avg := FunctionAverage( f(x), x=a..b, output=value ); 

evalf( avg ); 

FunctionAverage(f(x),x=a..b,output=plot); # (d) 

fsolve( f(x)=avg, x=0.5 ); 

fsolve( f(x)=avg, x=2.5 ); 

fsolve( f(x)=Avg[1000], x=0.5 ); 

fsolve( f(x)=Avg[1000], x=2.5 ); 
Mathematica : (assigned function and values for a and b may vary): 
Symbols for n, — > , powers, roots, fractions, etc. are available in Palettes (under File). 
Never insert a space between the name of a function and its argument. 

Clear[x] 

f[x_]:=x Sin[l/x] 

{a,b} = {7r/4, tt} 

Plot[f[x],{x, a, b}] 
The following code computes the value of the function for each interval midpoint and then finds the average. Each 
sequence of commands for a different value of n (number of subdivisions) should be placed in a separate cell. 

n=100;dx = (b-a)/n; 

values = Table[N[f[x]], {x, a + dx/2, b, dx}] 

average=Sum[values[[i]],{i, 1, Length[values]}] / n 

n =200; dx = (b - a) /n; 

values = Table[N[f[x]],{x, a + dx/2, b, dx}] 

average=Sum[values[[i]],{i, 1, Length[values]}] / n 

n=1000;dx = (b-a)/n; 

values = Table[N[f[x]],{x, a + dx/2, b, dx}] 

average=Sum[values[[i]],{i, 1, Length[values]j] / n 

FindRoot[f[x] == average, {x, a}] 

5.2 SIGMA NOTATION AND LIMITS OF FINITE SUMS 



1- E E 



6k 6(1) , 6(2) _ 6 , 12 

+ 1 1 + 1 """ 2+1 2^3 



z " ^ k 1 T 2 ^ 3 u ^ 2 ^ 3 6 



3. J2 cos kit = cos(l7r) + cos(27r) + cos(37r) + cos(47r) = —1 + 1 — 1 + 1=0 

k=l 



4. J2 sin k7r = sin (l7r) + sin (27r) + sin (37r) + sin (4tt) + sin (57r) = + + + + = 



5. £ (-l) k+1 sin I = (-1) 1+1 sin f + (~1) 2+1 sin f + (-1) 3+1 sin f = - 1 + ^ = A^ 



6. Y, (-1)" cos kn = (-1) 1 cos(Itt) + (-1) 2 cos(2tt) + (-1) 3 cos(3tt) + (-1) 4 cos(4?r) 

k=l 

= -(-!)+ !-(-!)+ 1 = 4 



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Section 5.2 Sigma Notation and Limits of Finite Sums 

-' - 2 1 - 1 + 2 2 - 1 + 2 3 - 1 + 2 4 - 1 + 2 5 - 1 + 2 6 - 1 

(b) ± 2 k = 2° + 2 1 + 2 2 + 2 3 + 2 4 + 2 5 = 1 + 2 + 4 + 

(c) E 2 k ^ 



299 



7. (a) E 2"- 1 =2' ' : 2 

k=l 



1 + 2 + 4 + 8+16 + 32 
+ 16 + 32 
2 -i 1 + 2 o+i + 2 1+1 + 2 2+1 + 2 3+1 + 2 4+1 = 1+2 + 4+8+16 + 32 



All of them represent 1+2 + 4 + 8+16 + 32 

(a) E (-2) k -' = (-2) 1 - 1 + (-2) 2 - 1 + (-2) 3 - 1 + (-2) 4 - 1 + (-2) 5 - 1 + (-2) 6 - 1 = 1-2 + 4-8+16-32 

k=l 

(b) J2 (-l) k 2 k = (-l)°2° + (-l) 1 2 1 + (-l) 2 2 2 + (-l) 3 2 3 + (-l) 4 2 4 + (-l) 5 2 5 =1-2 + 4-8+16-32 

k=() 

(c) J2 (-l) k+1 2 k + 2 = (-l)- 2+1 2- 2 + 2 + (-1)- 1+1 2- 1+2 + (-1) 0+1 2 0+2 + (-1) 1+1 2 1+2 + (-1) 2+1 2 2 + 2 

k=-2 

+ (-l) 3+1 2 3 + 2 = -1 +2-4 + 8- 16 + 32; 
(a) and (b) represent 1 — 2 + 4 — 8+16 — 32; (c) is not equivalent to the other two 



9- (a) E ^l 



+ 



(-1)* 



2-1 ^ 3-1 



1 + 



2 3 



k+1 0+l'l+l'2+l 2'3 



(b)E^ = ^ + ^-^ 



(c) E 



k + 2 -1 + 2 ' + 2 ' 1+2 



+ 



•1 + i - i 



(a) and (c) are equivalent; (b) is not equivalent to the other two. 

10. (a) E ( k - !) 2 = (1 - I) 2 + (2 - l) 2 + (3 - l) 2 + (4 - l) 2 = + 1 + 4 + 9 

k=l 

(b) E (k + I) 2 = (-1 + I) 2 + (0 + l) 2 + (1 + l) 2 + (2 + l) 2 + (3 + l) 2 = + 1 + 4 + 9 + 16 

k=-l 

(c) E k 2 = (-3) 2 + (-2) 2 + (-l) 2 = 9 + 4+l 

k=-3 

(a) and (c) are equivalent to each other; (b) is not equivalent to the other two. 



11. Ek 



12. E k 2 



13. E i 



14. E 2k 



15. E(-D k+1 i 



16. E(-D k I 



17. (a) E 3a k = 3 E a k = 3(-5) = -15 

k=l k=l 

(b) E h i = ltw = l(6)=l 

k=l k=l 

(c) E (a k + b t ) = E a t + E b t = -5 + 6 = 1 

k=l k=l k=l 

(d) E (a k - b t ) = E a t - E bk = -5 - 6 = -11 



k=[ k=l 



(e) E (bk - 2a k ) = Eb k -2Ea k = 6- 2(-5) = 16 

k=l k=l k=l 



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300 Chapter 5 Integration 

18. (a) £ 8a k = 8 E a k = 8(0) = 



n n 



(c) E (a* + 1) = E a K + E 1 = ° + n = n 

k=l k=l k=l 



(b) E 2 50b k = 250 E W = 250(1) = 250 



k=l 
n n 



(d) E(b k -D = Eb k -El = l-n 

k=l k=l k=l 



19. (a) Ek 



10(10+1) 



55 



(c) £ k 3 = [wQO+iij 2 = 552 = 3025 



(b) Ek : 



2 _ 10(10+ 1)(2(10)+1) 



385 



20. (a) Ek 



k=l 
13 



13(13-1) 

2 



91 



(c) J2 k 3 = [i32|±i2l = 91 s 



8281 



(b) Ek 



2 _ 13(13 + 1X2(13)+ 1) 



819 



21. E -2k = -2 E k = -2 f 71 ^ 12 ) 



-56 



22. 



E f5 = BEk=f3^)=- 

k=l k=l x ' 



1 , _ V^ 1 V^ v2 _ ^^^ 6(6 + l)(2(6)+l) 



23. E(3-k 2 )=E3-Ek 2 = 3(6) 



-73 



k=l k=l 



A <\ - SP V2 V^ c _ 6(6+l)(2(6)+l) 



24. £( k 2-5) = Ek 2 -E5 



5(6) = 61 



25. E k(3k + 5) = E (3k 2 + 5k) = 3 E k 2 + 5 E k = 3 ( 5(5 + 1)( 6 2(5)+1) ) + 5 f 5 ^) 

V — 1 1- — 1 V — 1 V — 1 ** / \ / 



240 



k=l k=l 



26. 



E k(2k + 1) = E (2k 2 + k) = 2Ek 2 + Ek = 2 ( 7(7 + 1)( 6 2(7)+1) ) 

1,-1 1.— 1 1-— i i — i V * 



7(7+1X2(7)+!)^ , 7(7+1) 



308 



k=l k=l 



27. g ^+(Ek) 3 = ^Ek3 + (E k y=^(^ 2 +(5^) 3 = 3376 



k=l / k=l 



28. Ek -Ef= Ek -iEk 3 = (^XH 



4 ^ =588 



29. (a) 



(b) 



(c) 



(2,3) 



0<x<2 

Left-hand 



c, =0 c ? c/= 1 c d 2 



a- 



2 L 3 , M 



(2,3) 



/(x)=x z -l, 
< .t < 2 
Right-hand 







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Section 5.2 Sigma Notation and Limits of Finite Sums 301 



30. (a) 



(b) 



(c) 



f(x)=-x z 






-1 



f(x) = -x' 



"■" crm 




f(x)=-x 



^J 



31. (a) 



(b) 



(c) 




f(x) = sin x, 

-ir<X<TT 

Right-hand 



h XTX. 





32. (a) 



2 
1.5 




Willi 




-Jt -K/2 



-0.5 

-1 



:■■:■:■:■:/.■;:.■;-:■:■:■■;. 



(b) 



71/2 it 

f(x) = sin x +1 



y 

2r .M.w. . .M ., 



1.5 




-jt -ji/2 



-0.5 
-1 



(c) 



Jl/2 It 

f(x) = sin x +1 



2 1 


f 




1.5 


/ 


111 


viy 


-Jt -rc/2 

-0.5 

-1 


Jt/2 7C 

f(x) - sin x +1 



33. |xj - x | = |1.2- 0| = 1.2, |x 2 - xi| = |1.5 - 1.2| = 0.3, |x 3 - x 2 | = |2.3 - 1.5| = 0.8, |x 4 - x 3 | = |2.6 - 2.3 1 = 0.3, 
and |x 5 - x 4 | = |3 - 2.6| = 0.4; the largest is ||P|| = 1.2. 

34. | Xl - x | = 1-1.6 - (-2)| = 0.4, |x 2 - xj| = | — 0.5 - (-1.6)| = 1.1, |x 3 - x 2 | = |0 - (-0.5)| = 0.5, 
|x 4 -x 3 | = |0.8-0| =0.8, and|x 5 - x 4 | = |1 - 0.8| = 0.2; the largest is ||P|| = 1.1. 




Since f is decreasing on [ 0, 1] we use left endpoints to obtain 
upper sums. Ax = ^— ^ = - and x; = iAx = -. So an upper sum 

is E(i-*?)H^S(i-(;o 2 ) = ^£(n 2 -i 2 ) 

i=0 i=0 v ' i=0 



^ 2 = i 

i=0 



i=0 x ' i=0 

(n-l)n(2(n-l) + l) _ 1 2n 3 -3n 2 +n 



6n 3 



1 



-^ — — . Thus, 

6 ' 



lim E(l-xf)i=Mm 1- 

n— >oo .ft n— t™ v 



l - 



Gn3 



12 
3 3 



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302 Chapter 5 Integration 




Since f is increasing on [ 0, 3] we use right endpoints to obtain 
upper sums. Ax = ^2 = 2 anc j Xj = [/\ x = 21 Soan U pp er 



sum is E2x 5 (|)= Ef -I = ^Ei 

i=l i=l i=l 



18 n(n + l) _ 9n 2 + 9n 
n 2 2 ~~ n 2 



Thus, lim E~ • " = lim^ 



9n 2 + 9n 

,2 



lim (9- 

n— >oo 



9. 




Since f is increasing on [ 0, 3] we use right endpoints to obtain 
upper sums. Ax = ^2 = 3 an( j x . _ j^ x _ & Soan upper 

sumisE(x? + l)l= E((!) 2 + l)! = |E(f + l 

i=l i=l v x i=l v ' 

= fEi 2 + f-n=g( n(n + 1) c (2n + 1) )+ 3 



i-l 

9(2n 3 + 3n 2 + n) 
2n 3 



o 18 +f + 5 , o Tk 

3 = 5 — =- + 3. Thus, 



lim E(x 2 + l)i=lim 



18+^ + - 



3 =9 + 3 = 12. 




Since f is increasing on [ 0, 1] we use right endpoints to obtain 
upper sums. Ax = -!— 2 — I and x; = iAx = L . Soan upper sum 



is 



X>? (I) = E3(^) 2 (D = I Ei 2 = Jr • ( n(n+ T n + 1) ) 

i=l i=l i=l v ' 



2n J + 3n 2 + n 



• '-•■" 2 ti_2. Thus, lim E3x, 2 (i) 

2 n^oo g ' Kn) 



2n 3 



= lim 

n^oo 



2+S + 



2 j 2 ± - 



39. f(x) = x + x 2 =x(l + x) 
y 




Since f is increasing on [ 0, 1] we use right endpoints to obtain 
upper sums. Ax = ^— ^ — I anc [ Xj — j/\ x — I So an upper sum 

i=l i=l v i=l i=l 

2n 3 + 3n 2 + n 



1 / n(n+l) \ 1 / n(n+l)(2n + l) \ _ n 2 + n , 

n 2 \^ 2 J + n l\ 6 y~ 2n 2 "+" 

n 

Thus, lim E( x i + x2 )s 



6n 3 



i±i 

2 



3,1 



= lim 

n^oo 



m 



n^oo 

2 + 2 + 4, 
6 



1 , 2 _ 5 

2 ' G G - 




Since f is increasing on [ 0, 1] we use right endpoints to obtain 
upper sums. Ax = -!— 2 — I and x; = iAx = -. Soan upper sum 

isE(3 X i + 2x 2 )i = E(f + 2(D 2 )n = ,li:i+4Ei 2 

i=l i=l v • i=l i=l 

3n 2 + 3n , 2n 2 + 3n + 1 



3 f n(n+l) \ 2 f n{n+ l){2n + 1) \ 

n 2 V 2 ) + n 3 ^ - 6 " ) 



2n 2 



3n 2 



2 + ; + -V 



-+ 3 " 2 . Thus, lim E(3x, + 2x 2 )i 



= lim 

n^oo 



m 



n^oo 

2 + n + S 

3 



3 j. 2 _ 13 
2^3 - 



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5.3 THE DEFINITE INTEGRAL 



Section 5.3 The Definite Integral 303 



1. fx 2 dx 

Jo 

4. f 1 dx 

J 1 x 

r° 

7. | (sec x) dx 

J -tt/4 



2. 



dx 



8. J (tan x) dx 



h i-x 



3. J\(x 2 -3x)dx 
6. J a/4 - x 2 dx 



9. (a 

(c 
(c 
(f) 

10. (a 

(b 
(c 
(d 
(e 

(f) 

11. (a 

(c 

12. (a 

(c 

13. (a 
(b 

14. (a 
(b 



/ g(x)dx = 



(b) 



f 5 g(x) dx = -f i g(x) dx = -8 



J 3f(x)dx = 3j f(x) dx = 3(-4) = -12 (d) J f(x) dx = J f(x) dx - J f(x) dx = 6 - (-4) = 10 

/ x [f(x) - g(x)] dx = £ f(x) dx - J x g(x) dx = 6 - 8 = -2 

Jj [4f(x) - g(x)] dx = 4 Jj f(x) dx - f t g(x) dx = 4(6) -8 = 16 

n9 p9 

J -2f(x)dx = -2J f(x)dx= -2(-l) = 2 

n9 p9 ^9 

J. [f(x) + h(x)] dx = J f(x) dx + J. h(x) dx = 5 + 4 = 9 

J 7 [2f(x) - 3h(x)] dx = 2 j 7 f(x) dx - 3 J 7 h(x) dx = 2(5) - 3(4) = -2 

f g f(x) dx = - J] f(x) dx = -(-1) = 1 

/ x f(x) dx = J] f(x) dx - J 7 f(x) dx = -1 - 5 = -6 

J* g [h(x) - f(x)] dx = J 7 [f(x) - h(x)] dx = j 7 f(x) dx - f 7 h(x) dx = 5 - 4 = 1 



J] f(u) du = J] f(x) dx = 5 

j 2 if « dt = - r f « dt = - 5 



(b) J x y/3 f(z) dz = a/3 / x f(z) dz = 5 a/3 
(d) J* [-f(x)] dx = - Jj f(x) dx = -5 



/; g (t)dt=-/_ 3 g(t)dt=-A/2 
f_ a [-g(x)] dx = - j 3 g(x) dx = -a/2 

/ a f(z) dz = f o f(z) dz - J fl f(z) dz = 7 - 3 = 4 
J 4 3 f(t)dt=-/ 3 4 f(t)dt=-4 



(b) J 3 g(u) du = J_ a g(t) dt = a/2 

»> r 3 f dr =v3r3^ dt =fe)(^)= 1 



p3 n3 r« 1 

J h(r) dr = J h(r) dr - J h(r) dr = 6 - = 6 
- J 3 h(u) du = - ( - J] h(u) du") = Jj h(u) du = 6 



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304 Chapter 5 Integration 

15. The area of the trapezoid is A = | (B + b)h 

= \ (5 + 2)(6) = 21 => J 2 (| +3) dx 
= 21 square units 




16. The area of the trapezoid is A = I (B + b)h 

P 3/2 

= i (3 + !)(!) = 2 



f 3/2 

J (-2x + 4)dx 



1/2 



2 square units 



17. The area of the semicircle is A 



2 " L 2 



i-(3) 2 



/-3 ^ - ^ 



dx 



7r square units 



18. The graph of the quarter circle is A 



|^(4) 2 



47r =>• I v 1 6 — x 2 dx = 47r square units 



■f f(x) - -2x + 4 




0,5 1 1,5 2 




-3 -2 -1 



12 3 




19. The area of the triangle on the left is A = \ bh = \ (2)(2) 
= 2. The area of the triangle on the right is A = \ bh 



|(1)(D 



Then, the total area is 2.5 



r 



|x| dx = 2.5 square units 




20. The area of the triangle is A = \ bh = \ (2)(1) = 1 
=>• I (1 — |x|) dx = 1 square unit 



Section 5.3 The Definite Integral 305 

y 

f(x) = 1-|x| 




21. The area of the triangular peak is A = \ bh = \ (2)(1) = 1. 
The area of the rectangular base is S = £w = (2)(1) = 2. 

Then the total area is 3 => I (2 — jx|) dx = 3 square units 




22. y = 1 + \f\ -X 2 => y -\ = y/\ -X 2 

=>■ (y - l) 2 = 1 - x 2 =>• x 2 + (y - l) 2 = 1, a circle with 
center (0, 1) and radius of 1 => y = 1 + y 1 — x 2 is the 
upper semicircle. The area of this semicircle is 
A = 1 7rr 2 = 1 7r(l) 2 = |. The area of the rectangular base 
is A = £w = (2)(1) = 2. Then the total area is 2 + | 



I ll + vl— x 2 )dx = 2+f square 



units 



) = 1 +Vi-> 




23. J b fd^=5(b)(^) 



1 /wt^ _ b^ 
4 




r b 

24. J 4x dx = \ b(4b) = 2b 2 



4b iV" 4 " 




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306 Chapter 5 Integration 

25. J 2s ds = \ b(2b) - \ a(2a) = b 2 - a 2 




y-2s 



26. J a 3t dt = \ b(3b) - \ a(3a) = § (b 2 - a 2 ) 





/ 




3b 




A) 

/ i 


3a 


/ 


i 

i 

i 
i 

■ 




a 


b 



,y = a 



27. f^d^-f = | 



/; 



J 1 



29. 5 d0 



(2tt) 2 _ ^ 3tt1 
2 2 2 



3,//WM H 



33. 



J'l/2 ,^3 

t 2 dt=if = i 



J. 2.") 
0.5 



28 . | 2 ; 5 xdx= (2|)!_(0|)! = 3 



B ^_M! Ml 2 



30. J,, rdr 



2 2 



24 



X 



0.3 

32. | s 2 ds = &P- = 0.009 



/W 2 m 3 t 

34. 6> 2 d6» — — - 

Jo 



3 — 24 



J. 2a 
a 



35. f xd x=^-f = f 



37. 



^b 



X 





dx 



39. J 7dx = 7(l-3) = -14 



41. J o 5xdx = 5 J o xdx = 5 [ 



2 2 2 



10 



43. 



44. 



/ o 2 (2t-3)dt = 2j r i 1 tdt-/ o 2 3dt = 2[ 



L 



36. I xdx=£$>--£=.» 



r 3b 

X 2 

Jo 



(3b) 3 



38. | x 2 dx = ^ = 9b 3 

'o 3 



40. 



J. -2 



dx= V2(-2-0)= -2\/2 



42 - fl\**=\fl***=\[\ 



1 | 51 31 _ 16 _ i 

2 2 16 - 1 



21 _ o! 

2 2 



3(2 - 0) = 4 - 6 = -2 



r 



t - V2 dt 



p\/2 P\/2 _ 

Jo tdt -Jo ^ dt 



fV£ 



-A/2U/2-0 = 1 -2= -1 



45. |„ (1 + 1) dz 



46. 



/ 2 1 ld Z+ / 2 1 ldz=/ 2 1 ldz-l/ i 2 zdz=l[l-2]-l[f-f] 



2 V2/< 4 



nO nO nO f.3 pO f , 

J 3 (2z - 3) dz = J 3 2z dz - J 3 3 dz = -2 J q z dz - J 3 3 dz = -2 If - f - 3[0 - 3] = -9 + 9 = 

47. /;3u 2 du = 3/>du = 3[£u 2 du-/\ 2 du]=3([f-f]-[f-f])=3[f-^=3(|)=7 

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Section 5.3 The Definite Integral 307 



/, 



i pi 

2 



f, U2 

J 1/2 



24u 2 du = 24 | u 2 du = 24 

/2 J 1/2 



pi pl/2 

I u 2 du — I u 

Jo Jo 



du 



24 



24 



(I) 



r<3 



x- + x - 5) dx = 3 / x 2 dx 



/ x 2 dx + / x dx I 

Jo Jo Jo 



xdx-| 5 dx = 3 I ^ - ^ 

o Jo 



f - f - 5[2 - 0] = (8 + 2) - 10 = 



50. 



J] (3x 2 + x - 5) dx = - J o (3x 2 + x - 5) dx = - 3 J Q x 2 dx + J Q x dx - J q 5 dx 

0r-¥) + 0r-S)-5(i-o)]=-(i-5) = ! 



51. Let Ax = ^ = ^ and let x = 0, Xi = Ax, 

n n u ' i ' 

X2 = 2Ax, . . . , x n _j = (n — l)Ax, x n = nAx = b. 

Let the c k 's be the right end-points of the subintervals 
=> ci = xi, C2 = X2, and so on. The rectangles 

defined have areas: 

f(ci) Ax = f(Ax) Ax = 3(Ax) 2 Ax = 3(Ax) 3 
f(c 2 ) Ax = f(2Ax) Ax = 3(2Ax) 2 Ax = 3(2) 2 (Ax) 3 
f(c 3 ) Ax = f(3Ax) Ax = 3(3Ax) 2 Ax = 3(3) 2 (Ax) 3 

f(c n ) Ax = f(nAx) Ax = 3(nAx) 2 Ax = 3(n) 2 (Ax) 3 
Then S n = ]T f(c k ) Ax = £ 3k 2 (Ax) 3 



3(Ax) 3 J2 k 

k= 



k=l 

3V 1-2-3 



b 3 



2 + L 



f\?\ / n(n+l)(2n+l) \ 

r 

Jo 



3x 2 dx = lim £(2+^ + 4, 

n ^ oo 2 V ' n ' n 2 



b 3 . 



3b 



f (x) = 3x 



(b,3b 



Xo-0 x, x 2 -x„_, x„.b 



52. Let Ax = ^_J> = k and let x = 0, Xi = Ax, 

X2 = 2Ax, . . . , x n _j = (n — l)Ax, x n = nAx = b. 
Let the c k 's be the right end-points of the subintervals 
=^> ci = xi, C2 = X2, and so on. The rectangles 
defined have areas: 

f(ci) Ax = f(Ax) Ax = tt(Ax) 2 Ax = tt(Ax) 3 
f(c 2 ) Ax = f(2Ax) Ax = tt(2Ax) 2 Ax = tt(2) 2 (Ax) 3 
f(c 3 ) Ax = f(3Ax) Ax = tt(3Ax) 2 Ax = tt(3) 2 (Ax) 3 

f(c n ) Ax = f(nAx) Ax = Tr(nAx) 2 Ax = 7r(n) 2 (Ax) 3 
Then S n = £ ffe) Ax = ]T 7rk 2 (Ax) 3 

k=l k=l 

= *(&*)'£ k' = ir(g)( » + l f + 11 ) 

Jo 



7Tb 3 



n + n^ 



Tb :; 



7tx 2 dx = lim 

n — » oc o 



'2+! + A) 



Ttb 



V 2 

(b.Jtb ) 



3 • 



f(x) = JtX 


2 

/ 






/ 




-"' 



X 

x. -b 



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308 Chapter 5 Integration 



53. Let Ax = ^ = b and let = Q = A 

n n u ' -l 

X2 = 2Ax, . . . , x n _! = (n — l)Ax, x n = nAx = b. 

Let the c k 's be the right end-points of the subintervals 
=> Ci = xi, C2 = X2, and so on. The rectangles 

defined have areas: 

f(ci) Ax = f(Ax) Ax = 2(Ax)(Ax) = 2(Ax) 2 
f(c 2 ) Ax = f(2Ax) Ax = 2(2Ax)(Ax) = 2(2)(Ax) 2 
f(c 3 ) Ax = f(3Ax) Ax = 2(3Ax)(Ax) = 2(3)(Ax) 2 

f(c n ) Ax = f(nAx) Ax = 2(nAx)(Ax) = 2(n)(Ax) 2 
Then S n = t %0 Ax = £ 2k(Ax) 2 



2(Ax) 2 Ek = 2(|)( n ^l) 



b 2 1 



k=l 
1 



Jo 



2x dx 



lim b 2 1 + 

n — > oo v 



2b 



v° 



1 

f(x) 


= 2x 

/ 


/ 


/ 



54. Let Ax = ^ = ^ and let x = 0, Xi = Ax, 

n n u ' i ) 

X2 = 2Ax, . . . , x n _j = (n — l)Ax, x n = nAx = b. 
Let the c k 's be the right end-points of the subintervals 
=^> ci = xi, C2 = X2, and so on. The rectangles 
defined have areas: 

f(ci) Ax = f(Ax) Ax = (^ + l) (Ax) = \ (Ax) 2 + Ax 
f(c 2 ) Ax = f(2Ax) Ax = (2^ + l) (Ax) = ^(2)(Ax) 2 + Ax 
f(c 3 ) Ax = f(3Ax) Ax = (2j* + l) (Ax) = \ (3)(Ax) 2 + Ax 




f(c) Ax = f(nAx) Ax = (sf* + l) (Ax) = \ (n)(Ax) 2 + Ax 
Then S n = t ffe) Ax = £ (± k(Ax) 2 + Ax) = \ (Ax) 2 tk + Axfl 

k=l k=l k=l k=l 



1 (b 2 

2 [ n^ 



ib 2 o + ^: 



1 dx 



lim i b 2 1 

n — > oo \4 \ 



b) 



|b 2 + b. 



n(n+l) 



(n) 



• Vs 



55. av(f) = (-^) J o \x 2 -l)dx 
x 2 dx - 4- I 1 dx 

\/3 Jo 

^ - (A = i - i = o. 



i 

v^ 







56. av(f)=(3^)/ o 3 (-f)dx=H-DX 3 

_ I f 3£^ - _3._x^_ _ 3 
6 I 3 J 2 ' 2 2 ' 



x 2 dx 




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57. av(f) = (^5) J o (-3x 2 - 1) dx = 

= - 3 / lx2dx -X lidx =- 3 (C)- (1 - 0) 



Section 5.3 The Definite Integral 



309 




58. av(f»=( T ^)/ o 1 (3x 2 -3) 



dx 



X lx2dx -X l3dx=3 (^)- 3(1 - 0) 




59. av(f)=(3^)/ o 3 (t-l) 2 dt 

n3 p3 r*6 

= Uo t2dt -IJo tdt +Uo 1 

= i(?)-i(f-f) + |(3-o> = i- 



dt 




0.5 1 1.5 2 15 3 



60. av(f)=( T ^)/_ 1 2 (t 2 -t) 



dt 



i/_>dt-i£tdt 



3 \tJ — 3 V - 3~ J ~*~ 2 



3 

2 2 




-2 -1.5 -1 -0.5 



61. 



- l)dx 



(a) av(g) = (jj-s) j\ (|x| - 1) 

= i/_° i (-x-l)dx+i/ o 1 (x-l)dx 

/0 nO nl nl 

-i« to -U-i ldx +Uo xdx -Uo 1 



dx 



l(f-^)-|(o-(-i)) + |(f-f)-|a-o) 



.. g(x)-M-i 



-1 

H — k—» 




■+—J 1 1 X 



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310 Chapter 5 Integration 



(b) av(g) = (3^) J x (|x| - 1) dx = \ Jj (x - 1) dx 

= iX 3 xdx-ij; 3 ldx=l(f-|)-i(3-l) 



(c) av(g) = (arfzij) /^(M - 1) dx 

= i/ 1 i (|x|-l)dx+i/ i 3 (|x|-l) 



dx 



7 (— 1 + 2) = i (see parts (a) and (b) above). 



H 1 1 H 



-!•■ 



■■ y-M-1 
■3 

• ■2 

■ 1 



-t 1 X 



12 3 



••2 



H h 



-1 



y-M-1 




12 3 



< 1 X 



62. (a) av(h) = (0^) J_ r |x| dx = f_ r (-x) dx 



x dx 



£ _ (-1) 2 _ _ 1 
2 2 2 ■ 



(b) av(h) 



= to)/o 1 -l« 

(V _ &\ _ _ 1 

^ 2 2 ^ 2' 



dx 



■/.' 

Jo 



x dx 



(c) av(h) = (^) £ - |x| c 

' /.O pi 

J-f-Mfc + Jo-l* 



= 2 (~ 2 + (~ 2)) = ~ 2 ( see P^ 8 W and (b) 
above). 



•■ 1 



-t — 1 — 1 — t- 



h(x)=-|x| 



-I 1 1 1 X 



■■-1 



■ ■ 1 



-i 1 1 H 



h(x) = -|x| 



H 1 1 X 



•■-1 



■■ 1 



■4 1 ► 



h(x)=-|x| 




I 1 1 1 X 



••-1 



63. To find where x - x 2 > 0, let x - x 2 = 
and b = 1 maximize the integral. 



=> x(l - x) = => x = Oorx = 1. If < x < 1, then < x - x 2 =^ a = 



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Section 5.3 The Definite Integral 311 



64. To find where x 4 - 2x 2 < 0, let x 4 - 2x 2 = => x 2 (x 2 - 2) = =>• x = or x = ± \fl. By the sign graph, 
++++++ +++++++, we can see that x 4 - 2x 2 < on l-y/l, \0\ =>a=-v/2andb 



V^ 



minimize the integral. 



65. f(x) = yt-3 is decreasing on [0, 1] =>■ maximum value of f occurs at =£• max f = f(0) = 1 ; minimum value of f 
occurs at 1 => min f = f(l) = jr^s = \ ■ Therefore, (1—0) min f < I jr-^ dx < (1 — 0) max f 



=> k < 



I y-Lj dx < 1 . That is, an upper bound = 1 and a lower bound 



66. See Exercise 65 above. On [0, 0.5], max f = y+& = 1. m in f = fxfea = 0.8. Therefore 

(0.5 - 0) min f < J Q f(x) dx < (0.5 - 0) max f => \ < J Q -^ dx < \ . On [0.5, 1], max f = l+ \ 05)2 = 0.8 and 



f = -r-^2 = 0.5. Therefore (1 - 0.5) min f < \ r-jU dx < (1 - 0.5) max f =>- \ < f y-f-j dx 

l + V v ' — Jo.5 l+ x — 4 — Jo.5 '+ x 

~n c ^ i _ i 



mm 

Then \ + 1 



< 



5 ' 



< 



n0.5 pi nl 

I T-U dx + I r-i-, dx < ± + f- =>• ^ < I -r-^ dx < 

JO 1 + x J 0.5 1 + x — 2 ' 5 20 — J o 1 + x — 



9_ 
10 ' 



67. -1 < sin(x 2 ) < 1 for all x => (1 -0)(-l) < J sin(x 2 )dx < (1 - 0)(1) or J sin x 2 dx < 1 =4> J 
equal 2. 



sin x 2 dx cannot 



68. f(x) = v/x + 8 is increasing on [0, 1] => max f = f(l) = y/i + 8 = 3 and min f = f(0) = v/O + 8 = 2\/l . 
Therefore, (1 - 0) min f < \ Jx + 8 dx < (1 - 0) max f => 2a/2 < f \/x + 8 dx < 3. 

*J t/ 

69. If f(x) > on [a, b], then min f > and max f > on [a, b]. Now, (b — a) min f < / f(x) dx < (b — a) max f. 
Then b > a => b - a > => (b - a) min f > => J f(x) dx > 0. 

70. If f(x) < on [a, b], then min f < and max f < 0. Now, (b - a) min f < J f(x) dx < (b - a) max f. Then 
b > a =4> b - a > =>• (b-a) max f < => J f(x) dx < 0. 

71. sin x < x for x > =4> sin x — x < for x > =>■ I (sin x — x) dx < (see Exercise 70) =>■ I sin x dx — I x 

< =4> I sin x dx < I x dx =>• I sin x dx < ( y — y J =4> I sin x dx < \ . Thus an upper bound is \ . 



dx 



72. sec x > 1 + \ on 



2 \ 2 ' 2 / 



sec x — 



(l + j)> 0cm (-!,!) =► /^[secx-^ + f^dx^O 
Exercise 69) since [0, 1] is contained in (— |, |) => I sec x dx — I (l + y)dx>0=> I sec x dx 



(see 



> 



£( 1 + l9 dx =* X SeCxdx ^£ ldx+ 2/ x2dx => X secxdx >( 1 -0)+l(T) =* J secxdx>|. 



Thus a lower bound is I. 



73. Yes, for the following reasons: av(f) = ^-^ I f(x) dx is a constant K. Thus / av(f) dx = I K dx 

r*b pb r»h 

= K(b - a) =4> J av(f) dx = (b - a)K = (b - a) • ^ J f(x) dx = J f(x) dx. 

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312 Chapter 5 Integration 



74. All three rules hold. The reasons: On any interval [a, b] on which f and g are integrable, we have: 

(a) av(f + g) = j^_ J a [f(x) + g(x)] dx = ^ | J] f(x) dx + X g(x) dx = ^-J & f(x) dx + ^ J] g(x) dx 

= av(f) + av(g) 

(b) av(kf) = g^ X kf W dx = b^i [ k J a f « dx ] = k [shi X f(x) dx = k av(f) 

r»b pb nb 

(c) av(f) = ^ J a f(x) dx < ^L J a g(x) dx since f(x) < g(x) on [a,b], and ^A J^ g(x) dx = av(g). 
Therefore, av(f) < av(g). 

75. Consider the partition P that subdivides the interval [a, b] into n subintervals of width Ax = ^^ and let c^ be the right 
endpoint of each subinterval. So the partition is P = {a, a + ^=^, a + ~ , . . . , a + ~ a) } and Ck = a H — ~ . 

We get the Riemann sum ^f(c k )Ax = JTc ■ ^ = ^^E 1 = £ ^ r il ■ n = c(b - a). As n -» oo and ||P|| -> 

k=l k=l ' k=l 

b 



this expression remains c(b — a). Thus, I c dx = c(b — a) 



76. Consider the partition P that subdivides the interval [a, b] into n subintervals of width Ax = - — - and let c^ be the right 
endpoint of each subinterval. So the partition is P = {a, a + ^F- 2 , a H — ( ~ , . . ., a + "^ ~ *' } and c^ = a + ~ ■ 



We get the Riemann sum £f(c k ) Ax = L>k(^) = l ^E(a+ k ^) 2 = ^E (a 2 + 



b-a f V^ Q 2 i 2a(b-a) 



^ E^+^pEk+^Ek 1 



k ! k=l k=l x ' k=l 



2ak(b - a) k 2 (b-a)' 



k-1 



k-1 



k-1 



(b-a^y^^l _ b-a „„2 i 2a(b-a)^ n(n+l) , (b-a) J n(n+l)(2n+l) 



n 



na 



n- 



+ 



ii- 



ii 



(b-a)a 2 + a(b-a) 2 .i±i + <V l! -^4 ±i 



= (b - a)a 2 + a(b - a) 2 • *±I + ^ • ™^' - "• ^~ 2 
As n — » oo and ||P|| — > this expression has value (b — a)a 2 + a(b — a) • 1 + * ~ a) • 2 

= ba 2 - a 3 + ab 2 - 2a 2 b + a 3 + i(b 3 - 3b 2 a + 3ba 2 - a 3 ) = f - f . Thus, J a x 2 dx = f - f . 

77. (a) U = maxi Ax + max2 Ax + . . . + max n Ax where maxi = f(xi), max2 = f(x2), . . . , max n = f(x n ) since f is 

increasing on [a, b]; L = mini Ax + min2 Ax + . . . + min n Ax where mini = f(xo), min2 = f(xi), . . . , 

min n = f(x n _i) since f is increasing on [a, b]. Therefore 

U — L = (maxi — mini) Ax + (max2 — min2) Ax + . . . + (max„ — min„) Ax 

= (f(xi) - f(x )) Ax + (f(x 2 ) - f(xi))Ax + . . . + (f(x n ) - f(x n _ { )) Ax = (f(x n ) - f(x )) Ax = (f(b) - f(a)) Ax. 
(b) U = maxi Axi -I- max 2 Ax2 + . . . + max„ Ax n where maxi = f(xi), max2 = ffe), . . . , max n = f(x n ) since f 

is increasing on[a, b]; L = mini Axi + min2 Ax 2 + . . . + min„ Ax„ where 

mini = f(xo), min 2 = f(xi), . . . , min n = f(x n _i) since f is increasing on [a, b]. Therefore 

U — L = (maxi — mini) Axi + (max2 — min2) Ax2 + . . . + (max n — min n ) Ax n 
= (f(xi) - f(x )) Axi + (f(x 2 ) - f(xj))Ax 2 + . . . + (f(x n ) - f(x n _,)) Ax„ 
< (f(xi) - f(x )) Ax raM + (f(x 2 ) - f( Xl )) Ax max + . . . + (f(x n ) - f(x n _,)) Ax max . Then 

U - L < (f(x n ) - f(x )) Ax max = (f(b) - f(a)) Ax max = |f(b) - f(a)| Ax max since f(b) > f(a). Thus 

lim (U - L) = lim (f(b) - f(a)) Ax mK = 0, since Ax max = ||P|| . 
||P|| -► ||P|| - " " 



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Section 5.3 The Definite Integral 313 



+ max n Ax where 
max n = f(x n _!) 



+• min n Ax where 
. min n = f(x n ) 




78. (a) U = maxi Ax + max2 Ax + . 
maxi = f(xo), max2 = f(xi), . 
since f is decreasing on [a, b]; 
L = mini Ax + min2 Ax + . . 
mini = f(xx), min 2 = f(x 2 ), . . 
since f is decreasing on [a, b]. Therefore 
U — L = (maxi — mini) Ax + (max 2 — min 2 ) Ax 

+ . . . + (max n — min n ) Ax 

= (f(x ) - f( Xl )) Ax + (f( Xl ) - f(x 2 ))Ax 

+ . . . + (ffr^) - f(x n )) Ax = (f(x ) - f(x n )) Ax 

= (f(a)-f(b))Ax. 
(b) U = maxi Axi + max 2 Ax 2 + . . . + max,, Ax n where maxi = f(xo), max 2 = f(xi), . . . , max n = ftx,,^) since 
f is decreasing on[a, b]; L = mini Ax x + min 2 Ax 2 + . . . + min,, Ax„ where 
mini = f( x i), min 2 = f(x 2 ), . . . , min n = f(x n ) since f is decreasing on [a, b]. Therefore 
U — L = (maxi — mini) Axi + (max 2 — min 2 ) Ax 2 + . . . + (max n — min„) Ax n 

= (f(x ) - f(xi)) Axi + (f(xi) - f(x 2 ))Ax 2 + . . . + (fCx^) - f(x„)) Ax n 

< (f(x ) - f(x n )) Ax mM = (f(a) - f(b) Ax max = |f(b) - f(a)| Ax,„ M since f(b) < f(a). Thus 

0, since Ax m „ = ||P|| . 



lim (U - L) 

||P||-0 



lim o |f(b) - f(a)| Ax„ 



79. (a) Partition [0, |] into n subintervals, each of length Ax = |j with points Xo = 0, Xi = Ax, 

x 2 = 2 Ax, ... , x„ = nAx = |. Since sin x is increasing on [0, |] , the upper sum U is the sum of the areas 
of the circumscribed rectangles of areas f(xi) Ax = (sin Ax)Ax, f(x 2 ) Ax = (sin 2Ax) Ax, . . . , f(x„) Ax 



(sin nAx) Ax. Then U = (sin Ax + sin 2Ax 



sin nAx) Ax 



cos 4f -cos( (n+ 1) Ax) 



2 sin ss 



Ax 



COS^- COS (( n+ l) jL) 



2 sin | 

4] 
n/2 



7r(cos^-cos(| + ^)) _ COS X_ cos (| + jL) 



(b) The area is I sin x dx 



r 

Jo 



lim 



cos f — cos 



4n sin -f- 

4n 
(f + fe) 



4n / 



m 



80. (a) The area of the shaded region is J2^ x i ' m i which is equal to L. 

i=l 

n 

(b) The area of the shaded region is X]^ x i ' Mj which is equal to U. 

i=l 

(c) The area of the shaded region is the difference in the areas of the shaded regions shown in the second part of the figure 
and the first part of the figure. Thus this area is U — L. 

n n 

81. By Exercise 80, U — L = J]^ x i ' Mj — J]^ x i ' m i where Mj = max{f(x) on the ith subinterval} and 

i=i i=i 

n n 

mi = min{f(x) on the ith subinterval}. Thus U — L = ^fMj — m;)Ax; < J^e ■ Axj provided Ax; < 6 for each 

i=l i=l 

n n 

i = 1, . . . , n. Since J2 e ' ^ x i = e S^ x i = e 3 ~ a ) m e result, U — L < e(b — a) follows. 

i=l i=l 



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314 Chapter 5 Integration 



82. The car drove the first 150 miles in 5 hours and the 
second 150 miles in 3 hours, which means it drove 300 
miles in 8 hours, for an average of ^9 mi/hr 
= 37.5 mi/hr. In terms of average values of functions, 
the function whose average value we seek is 
< t < 5 
5< 1 < 

(30)(5) + (50)(3) = 37 5 



Velocity 

mi/hr 
v 

501 



V(t) 



f 30, 
\50, 



30 



and the average value is 



_average 
"value 



37.5 mi/hr 



-t Time 
hr 



83-88. Example CAS commands: 
Maple : 

with( plots ); 

with( Student[Calculusl] ); 

f :=x -> 1-x; 

a:=0; 

b:=l; 

N:=[4, 10,20,50]; 

P := [seq( RiemannSum( f(x), x=a..b, partition=n, method=random, output=plot ), n=N )]: 

display ( P, insequence=true ); 

89-92. Example CAS commands: 
Maple : 

with( Student[Calculusl] ); 

f := x -> sin(x); 

a:=0; 

b := Pi; 

plot( f(x), x=a..b, title="#23(a) (Section 5.1)" ); 

N:=[ 100, 200, 1000]; # (b) 

for n in N do 

Xlist := [ a+l.*(b-a)/n*i $ i=0..n ]; 

Ylist := map( f, Xlist ); 
end do: 
for n in N do # (c) 

Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist)); 
end do; 

avg := FunctionAverage( f(x), x=a..b, output=value ); 
evalf( avg ); 

FunctionAverage(f(x),x=a..b,output=plot); # (d) 
fsolve( f(x)=avg, x=0.5 ); 
fsolve( f(x)=avg, x=2.5 ); 
fsolve( f(x)=Avg[1000], x=0.5 ); 
fsolve( f(x)=Avg[1000], x=2.5 ); 



83-92. Example CAS commands: 

Mathematica : (assigned function and values for a, b, and n may vary) 

Sums of rectangles evaluated at left-hand endpoints can be represented and evaluated by this set of commands 
Clear[x, f, a, b, n] 



Copyright (c) 2006 Pearson Education 




Section 5 .4 The Fundamental Theorem of Calculus 315 

{a, b}={0, tt}; n =10; dx = (b - a)/n; 

f=Sin[x] 2 ; 

xvals =Table[N[x], {x, a, b — dx, dx}]; 

yvals = f/.x — ► xvals; 

boxes = MapThread[Line[{{#l,0},{#l, #3}, {#2, #3), {#2, 0}]&,{xvals, xvals + dx, yvals}]; 

Plot[f, {x, a, b}, Epilog — ► boxes]; 

Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N 
Sums of rectangles evaluated at right-hand endpoints can be represented and evaluated by this set of commands. 

Clear[x, f, a, b, n] 

{a, b}={0, tt}; n =10; dx = (b - a)/n; 

f=Sin[x] 2 ; 

xvals =Table[N[x], {x, a + dx, b, dx}]; 

yvals = f /.x — > xvals; 

boxes = MapThread[Line[{{#l,0},{#l, #3}, {#2, #3), {#2, 0}]&,{xvals - dx,xvals, yvals}]; 

Plot[f, {x, a, b}, Epilog — » boxes]; 

Sum[yvals[[i]] dx, {i, l,Length[yvals]}]//N 
Sums of rectangles evaluated at midpoints can be represented and evaluated by this set of commands. 

Clear[x, f, a, b, n] 

{a, b}={0, tt}; n =10; dx = (b - a)/n; 

f=Sin[x] 2 ; 

xvals =Table[N[x], {x, a + dx/2, b - dx/2, dx}]; 

yvals = f /.x — > xvals; 

boxes = MapThread[Line[{{#l,0},{#l, #3}, {#2, #3}, {#2, 0}]&,{xvals - dx/2, xvals + dx/2, yvals}]; 

Plot[f, {x, a, b},Epilog — > boxes]; 

Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N 



5.4 THE FUNDAMENTAL THEOREM OF CALCULUS 



133 
4 



1. J 2 (2x + 5) dx = [x 2 + 5x]! 2 = (0 2 + 5(0)) - ((-2) 2 + 5(-2)) = 6 

2- £ (5 - 1) dx = [5x - f ] ^ = (5(4) - f ) - (5(-3) - ^) 

3. r(3x-f)dx=[f-fl];=(^-il)-(^-f)=8 

4. fjx 3 - 2x + 3) dx = [£ - x 2 + 3x] [ = (f - 2 2 + 3(2)) - (^ - (-2) 2 + 3(-2)) 



12 



;i + D-o = i 



5. X'( X 2 + V^)dx=[f + Ix 3 / 2 ]; 

6. J o 5 x 3 / 2 dx = [2 x 5 / 2 ] I = | (5) 5 / 2 - = 2(5) 3 / 2 = 10^ 

7. J" x-^ dx = [-5X-V5] » = (_§)_ (-5) = 1 

8 - £'* dx = £' 2x " dx = t- 2x "] = 2 = (3) - (-2) = 1 



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316 Chapter 5 Integration 

9. J sinxdx= [-cosx]^ = (-costt)-(-cosO) = -(-1)- (-1) = 2 

nir 

10. 1(1+ cos x) dx = [x + sin x]J = (n + sin 7r) — (0 + sin 0) = 7r 

11. J 2sec 2 xdx = [2tanx]g /3 = (2tan(f)) -(2tan0) = 2 v / 3-0 = 2 v / 3 



12. 



J csc 2 xdx = [-cotx] 5 ^ = (-cot(f )) - (-cot(f)) = - (-V3) - (-73) = 2^3 



2=0 



13. J esc cot d0 = [-esc 0]^ = (-csc(^)) - (-esc (f)) = -^2- 

14. J 4 sec u tan u du = [4 sec u]o /3 = 4 sec (f ) - 4 sec = 4(2) - 4(1) = 4 

15 f ktsa* dt= f (1 + i cos2t) dt= [lt+ 1 sin2t]° /2 = (1(0)+ i sin 2(0)) - (1 (|) + \ sin 2(f)) 



16. 



£l^f^ dt = £«(5 - I cos 2t ) dt = [I t - \ sin 2t] ^ 



; ^(I-|cos2t)dt=[it-isin2t] ; 
(I (f ) - ^sm 2 (f )) - (I (_ f ) - 1 sin 2 (- f )) 



= - i sin ^ + I + i sin (=^ - = - ^ 



6 4 "" 3 ' 6 ' 4 Ji " V 3 y 3 4 



17. £^ /2 (8y 2 + sin y) dy = [ 



7I-/2 



8y 3 " M(f) 3 ,A /8(-l) 3 

-r - cos y\ _ w/2 = (3 - cos I ) - I -V 1 - 



18. 



7T/4 

,< 



(4 sec 2 1 + 7rt~ 2 ) dt = [4 tan t - j] '^L 



19. 



20. 



X-7T/ 
-ir/3 

= (4 tan (- I) - ^) - (4 tan ( f ) - ^) = (4(-l) + 4) - (4 (-^3) + 3) = 4^3 - 3 
Jj (r+l) 2 dr=J i (r 2 + 2r+l)dr= [^+r 2 + r] _1 = (^£ + (-l) 2 + (-l)) - (| + l 2 +l) 



-VI 



/ (t + 1) (t 2 + 4) dt = J_^(t 3 + t 2 + 4t + 4) dt = [I + f + 2t 2 + 

= (l^ + l^ + 2(y^ + v^ 



16 



w 



2i -^-^) d -^- u i d -[^^]^(^w)-( 



.0* - & dv = /a 3 - v ") dv = [§a + 3^] u = (2W + w) - (# + 3#) 



22. 



23. //^ ds = J^(l + s-3/ 2 ) ds = [s - X] ^ = i ,/2 > 



1 - -4- ) = \/2 - 2 3 / 4 + 1 



^ - Vs + 1 



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Section 5 .4 The Fundamental Theorem of Calculus 317 

XV 1/2 - 1) du = [2^ - u] I = (2^4 - 4) - (2^9 - 9) = 3 

25j;jxid X =/;jxidx + /;ixidx=-/\dx + /;xdx = [-i]°_ 4+ [f]^(-f + ^) + (f-f) 



24 



r 



xA 



du 



16 



26. 



I | (cos x + |cos x| ) dx = I |(cos x + cos x) dx + I | (cos x — cos x) dx = I cos x dx = [sin x]g 



sin I — sin = 1 



Jo 



.V* 



^ => 4 f x 



27. (a) I cos t dt = [sin t]j = sin ^/x - sin = sin y/x => ^ ( I cos t dt ) = ^ (sin a/x) = cos ^x (| x~ 1/2 ) 



v^ 



2^1 
(b) ;t (X ™ Sldl 



7* 



(cos v^) (^ (v^)) = (cos ^x) (I x-V 2 ) = ^ 



28. (a) J 3t 2 dt= [t 3 ]i° x = sin 3 x- 1 ^ ^ (X 3t 2 dt ] = ^ (sin 3 x - 1) = 3 sin 2 x cos x 



(b) i ( X 3t 2 dt J = (3 sin 2 x) (£ (sin x)) = 3 sin 2 x cos x 



29. (a) f^du=f\^du=[j^Y i = j(t*f 2 -0=l? => |(/ V 



du I = |(|t 6 ) = 4t 5 



(b) 



X v^ du) = 7? (| (t 4 )) = t 2 (4t 3 ) = 4t 5 



30. (a) J sec 2 y dy = [tany]™ 9 = tan (tan 0) -0 = tan (tan 0) =$.±11 sec 2 y dy 1= ± (tan (tan 0)) 



(sec 2 (tan 0)) sec 2 



/ n lan \ 

(b) ± (J sec 2 y dy = (sec 2 (tan 9)) (± (tan 9)) = (sec 2 (tan 0)) sec 2 



31. y = / Vi+T 



2 dt=^ £ = v/r 



32 -y = fi 



dt => ^ = I , x > 

dx x ' 



33. y = I sint 2 dt= - / 



A 



sin t 2 dt 



2 Ht -^ ^y 



(^(v^) 2 ) (£ (v^)) = -(sinx)(|x-V2) = _^ 



34. y = X cos dt => I = (cos a/x 2 ) (5 (x 2 )) = 2x cos |x 



35. y 



/. 



M < f =-> £ = l . 1 .„ (£(sinx)) = -* (cosx) 



x/T^ ' ' ' 2 dx ~ x/l-sin^x Vdx 



««x = co^x = j since II < 1 

COS X COS X 112 



Jo 



36. y = .l. TT? =* Mli;)(l("")) = (iH)(»'') = ' 



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318 Chapter 5 Integration 

37. -x 2 - 2x = => -x(x + 2) = =4> x = or x = -2; Area 

= -J (-x 2 - 2x)dx + J_ 9 (-x 2 - 2x)dx - J (-x 2 - 2x)dx 



((- 



_ (- 2 ) 3 
3 

°?-0 2 

3 u 



(-2) 2 )-(-^-(-3) 2 )) 



(-2) 3 



(-2)' 



-^-2 2 W-f-0 2 ))=f 









/ 






H3 A2 

\ / 


-l 


-2 




1 




y = -x 2 - 2x 




-4 
-6 
-8 






\ J 
\ i 



38. 3x 2 — 3 = =^ x 2 = l =>- x= ±1; because of symmetry about 
they-axis, Area = 2 ( J (3x 2 - 3)dx + J (3x 2 - 3)dx 

2 (- [x 3 - 3x] I + [x 3 - 3x] I) = 2 [- ((l 3 - 3(1)) - (0 3 - 3(0))) 
+ ((2 3 - 3(2)) - (l 3 - 3(1))] = 2(6) = 12 




y = 3x -3 



39. x 3 - 3x 2 + 2x = => x (x 2 - 3x + 2) = 
=> x(x - 2)(x - D = => x = 0, 1, or 2; 

J (x 3 - 3x 2 + 2x)dx - J (x 3 - 3x 2 + 2x)dx 



Area 



\ -x 3 + x 2 - K -x 3 + x 2 

(^-i 3 + i 2 )-(^-o 3 + o 2 ) 

(^-2 3 +2 2 )-(^-l 3 +l 2 )]=I 



y - x - 3x + 2x 




40. x 3 - 4x = => x (x 2 - 4) = =>■ x(x - 2)(x + 2) = 

/0 n2 

(x 3 - 4x)dx - I (x 3 - 4x)dx 



2x< 



2x< 



o 1 



2(0) 2 ) 



(tfi - 2(-2) 2 ) - [{* - 2(2) 2 ) - (* - 2(0) 2 )] 




41. x 1 / 3 =0 =^ x = 0;Area= - J x 1 / 3 dx + J x 1 / 3 dx 

= [_3 x 4/3]0 + [3 x 4/3j 8 

= (~ I (0) 4/3 )- (" 1 (-1°) 4/3 ) + Q (8) 4 / 3 ) - (1 (O) 4 / 3 ) 



51 

4 



Copyright (c) 2006 Pearson Education, Inc 




Section 5.4 The Fundamental Theorem of Calculus 319 



42. x 1 / 3 - x = => x 1 / 


3 (l-x 2 / 3 ) -- 


= => 


xV 3 = 


Oor 


1 - x 2 / 3 = => x = 


= or 1 = x 2 / 3 => ; 


t = or 




1 = x 2 => x = or 


± i; 








Area = - 1 (x 1 / 3 


- x ) dx +r 


( x l/3 . 


- x) dx - 


r 


= -[l- 4/3 -f]° 


i + [|x 4 / 3 - 


2 Jo 


-[§x« 


_ x^ 


= - [(l(0) 4 / 3 -f 


)-(K-D 4/3 - k 


■n 




+ [(!d) 4/3 -f) 


-(|(0) 4 / 3 - 


?)] 






-[(!(8) 4/3 -f) 


-(i(D 4/3 - 


«)] 






= I + |-(-20- 


3 , 1\ 83 

4 ' 2) ~ 4 









.1/3 



>c)dx 



y 


1/3 

y = x -x 








\2 4 


6 8 


-l 






| 


-2 






| 


-3 








-4 






>. | 


-5 






>v 1 


-6 






N 



43. The area of the rectangle bounded by the lines y = 2, y = 0, x = ir, and x = is 27r. The area under the curve 



y = 1 + cos x on 



[o^iis/; 



(1 + cos x) dx = [x + sin x][J = (ir + sin ir) — (0 + sin 0) = it. Therefore the area of 



the shaded region is 2tt — it = ir. 
44. The area of the rectangle bounded by the lines x=|,x=^,y = sin| = i = sin^, and y = is 



1 (5n _ 7T\ _ 7T 

2 V 6 6) 3 



n 57r/6 
ir/6 



. The area under the curve y = sin x on [|, ^] is I sin x dx = [—cos x]J, 
(—cos ^) — (—cos |) = — (— 2i^~2 = V 3' Therefore the area of the shaded region is \J 3 — | . 



45. On [- f ,0] : The area of the rectangle bounded by the lines y = \fl, y = 0, 9 = 0, and 9 = - f is \fl (f ) 



-7T/4 



= ^(— . The area between the curve y = sec 6 tan 6 and y = is — I sec 8 tan A9 = [—sec 9]° 

= (—sec 0) — (—sec (— f )) = v2 — 1. Therefore the area of the shaded region on [— |,0] is ^ — h ( v 2 — 1 j 

On [0, f ] : The area of the rectangle bounded by 8 = J , 9 = 0, y = \fl, andy = is \fl (f ) = ^ . The area 

under the curve y = sec 9 tan 9 is I sec 8 tan 9 dd = [sec #]q = sec | — sec = y 2 — 1 . Therefore the area 

of the shaded region on [0, j\ is ^j 

(^ + V / 2-l) + (^- V / 2+l) 



2 — 1 ) . Thus, the area of the total shaded region is 



7IV2 



46. The area of the rectangle bounded by the lines y = 2, y = 0, t = — ^, and t = 1 is 2 (l — (— J) J = 2+ f . The 
area under the curve y = sec 2 1 on [— |, 0] is I sec 2 1 dt = [tan t]^_, 4 = tan — tan (— |) = 1. The area 



under the curve y = 1 - t 2 on [0, 1] is J ( 1 - t 2 ) dt = 1 1 - |1 = M - y J 



area under the curves on [—J, l] is 1 



Therefore the area of the shaded region is (2 



- V ) - ( - f ) = I . Thus, the total 



5 _ 1 , 7T 

3 ~~ 3 t 2 



47. y = J i dt - 3 =4> ^ = \ and y(?r) = J \ dt - 3 = - 3 = -3 =4> (d) is a solution to this problem. 

48. y = J sectdt + 4 => ^|= sec x and y(-l) = J sec t dt + 4 = + 4 = 4 =4> (c) is a solution to this problem. 

49. y = I sec t dt + 4 => ^ = sec x and y(0) = I sec t dt + 4 = + 4 = 4 => (b) is a solution to this problem. 

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320 Chapter 5 Integration 

50. y = J 1 dt - 3 => ^ = I and y(l) = J i dt - 3 = - 3 = -3 => (a) is a solution to this problem 

51. y = J sectdt + 3 52. y= J v 7 ! + t 2 dt - 2 
53. s = / f(x) dx + s 54. v = I g(x) dx + v 

" to *> <0 

55. Area=J^(h-(g)x 2 )dx=[hx """^ 



4hx 3 
3b 2 



b\ 4h(|)- 



2 y 3b 2 



b\ 4h(-|)' 



2/ 3b 2 



-bh w^ (_bh + bhx bh _bh = 2 bh 



2 6 



3 — 3 



h y = h - (4h/b 2 )x* 




-b/2 



56 - r =X( 2 -^ 



) dx = 2 r( i -(^w) dx = 2 [ x -(^T)]o= 2 [( 3+ (^))-(° + ^)] 



2 [3 \ - 1] = 2 (2 i) = 4.5 or $4500 



57- | = 5$5 = I ^ 1/2 =* ^ = Jj t^^dt = [tV] ; = ^ 



c(100) - c(l) = yiOO -\[\ = $9.00 



58. By Exercise 57, c(400) - c(100) 



'400 



100 = 20- 10 = $10.00 



59. (a) v = | = £ J f(x) dx = f(t) =>• v(5) = f(5) = 2 m/sec 

(b) a = ^j is negative since the slope of the tangent line at t = 5 is negative 

(c) s = I f(x) dx = | (3)(3) = | m since the integral is the area of the triangle formed by y = f(x), the x-axis, 

and x = 3 

(d) t = 6 since from t = 6 to t = 9, the region lies below the x-axis 

(e) At t = 4 and t = 7, since there are horizontal tangents there 

(f) Toward the origin between t = 6 and t = 9 since the velocity is negative on this interval. Away from the 
origin between t = and t = 6 since the velocity is positive there. 

(g) Right or positive side, because the integral of f from to 9 is positive, there being more area above the 
x-axis than below it. 

60. (a) v = f = | J q g(x) dx = g(t) => v(3) = g(3) = m/sec. 

(b) a = ^ is positive, since the slope of the tangent line at t = 3 is positive 

(c) At t = 3, the particle's position is J g(x) dx = | (3)(-6) = -9 

f 6 

(d) The particle passes through the origin at t = 6 because s(6) = | g(x) dx = 

(e) At t =7, since there is a horizontal tangent there 

(f) The particle starts at the origin and moves away to the left for < t < 3. It moves back toward the origin 
for 3 < t < 6, passes through the origin at t = 6, and moves away to the right for t > 6. 

(g) Right side, since its position at t = 9 is positive, there being more area above the x-axis than below it at t = 

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Section 5.4 The Fundamental Theorem of Calculus 321 



61. k > => one arch of y = sin kx will occur over the interval [0, |] =>- the area = I sin kx dx = [— j cos kx] 



7r/k 



-E cos ( k (!))-(-E cos (0)) 



62 - SS?/o"A 



dt = lim 



/o^ dt 



lim-M**- = lim^T-A 



x^O 



x^0° 



oc. 



63. 



64. 



J f(t) dt = x 2 - 2x + 1 => f(x) = £ J f(t) dt = £ (x 2 - 2x + 1) = 2x - 2 
I f(t) dt = x cos 7rx =>■ f(x) = gj I f(t) dt = cos 7rx — 7rx sin 7rx => f(4) = cos 7r(4) — 7r(4) sin 7T(4) = 1 



J'X+1 /•!+! 

^dt => f' W = _ TT ^ = -zL =, f'(l) = -3;f(l) = 2-J ^=2-0 = 2; 



I 2 i "I- i 1 + (x + 1) 

L(x) = -3(x - 1) + f(l) = -3(x - 1) + 2 = -3x + 5 

r*x 2 

66. g(x) = 3 + J sec(t- 1) dt => g'(x) = (sec (x 2 - l))(2x) = 2x sec (x 2 - 1) => g'(-l) = 2(-l) sec((-l) 2 - 1) 

r (-i) 2 pi 

= -2;g(-l)= 3 + J sec(t- 1) dt = 3 + J sec(t- 1) dt = 3 + = 3; L(x) = -2(x- (-l)) + g(-l) 

= -2(x+ l) + 3 = -2x+ 1 

67. (a) True: since f is continuous, g is differentiable by Part 1 of the Fundamental Theorem of Calculus. 

(b) True: g is continuous because it is differentiable. 

(c) True, since g'(l) = f(l) = 0. 

(d) False, since g"(l) = f'(l) > 0. 

(e) True, since g'(l) = and g"(l) = f'(l) > 0. 

(f) False: g"(x) = f'(x) > 0, so g" never changes sign. 

(g) True, since g'(l) = f(l) = and g'(x) = f(x) is an increasing function of x (because f'(x) > 0). 

68. (a) True: by Part 1 of the Fundamental Theorem of Calculus, h'(x) = f(x). Since f is differentiable for all x, 

h has a second derivative for all x. 

(b) True: they are continuous because they are differentiable. 

(c) True, since h'(l) = f(l) = 0. 

(d) True, since h'(l) = and h"(l) = f'(l) < 0. 

(e) False, since h"(l) = f'(l) < 0. 

(f) False, since h"(x) = f'(x) < never changes sign. 

(g) True, since h'(l) = f(l) = and h'(x) = f(x) is a decreasing function of x (because f'(x) < 0). 



69. 



= cosx 




70. The limit is 3x 2 
-h-0.5 




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322 Chapter 5 Integration 

71-74. Example CAS commands: 
Maple : 

with( plots ); 

f := x -> x A 3-4*x A 2+3*x; 

a:=0; 

b:=4; 

F := unapply( int(f(t),t=a..x), x ); # (a) 

pi := plot( [f(x),F(x)], x=a..b, legend=["y = f(x)","y = F(x)"], title="#71(a) (Section 5.4)" ): 

pi; 

dF := D(F); # (b) 

ql := solve( dF(x)=0, x ); 

ptsl := [ seq( [x,f(x)], x=remove(has,evalf([ql]),I) ) ]; 

p2 := plot( ptsl, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where F '( x )=0" ): 

display( [pl,p2], title="71(b) (Section 5.4)" ); 

incr := solve( dF(x)>0, x ); # (c) 

deer := solve( dF(x)<0, x ); 

df := D(f); # (d) 

p3 := plot( [df(x),F(x)], x=a..b, legend=["y = f '(x)","y = F(x)"], title="#71(d) (Section 5.4)" ): 

p3; 

q2 := solve( df(x)=0, x ); 

pts2 := [ seq( [x,F(x)], x=remove(has,evalf([q2]),I) ) ]; 

p4 := plot( pts2, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where f '00=0" ): 

display( [p3,p4], title="71(d) (Section 5.4)" ); 

75-78. Example CAS commands: 
Maple : 
a:= 1; 

u := x -> x A 2; 
f :=x->sqrt(l-x A 2); 
F := unapply( int( f(t), t=a..u(x) ), x ); 
dF := D(F); # (b) 

cp := solve( dF(x)=0, x ); 
solve( dF(x)>0, x ); 
solve( dF(x)<0, x ); 
d2F := D(dF); # (c) 

solve( d2F(x)=0, x ); 
plot( F(x), x=-l..l, title="#75(d) (Section 5.4)" ); 

79. Example CAS commands: 
Maple : 

f:=T; 

ql := Diff( Int( f(t), t=a..u(x) ), x ); 

dl := value( ql ); 

80. Example CAS commands: 
Maple : 

f:=T; 

q2 := Diff( Int( f(t), t=a..u(x) ), x,x ); 



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Section 5.5 Indefinite Integrals and the Substitution Rule 323 



value( q2 ); 



71-80. Example CAS commands: 

Mathematica l (assigned function and values for a, and b may vary) 

For transcendental functions the FindRoot is needed instead of the Solve command. 

The Map command executes FindRoot over a set of initial guesses 

Initial guesses will vary as the functions vary. 

Clear[x, f, F] 

{a, b}= {0, 2tt}; f[x_] = Sin[2x] Cos[x/3] 

F[x_] = Integrate [f[t], {t, a, x}] 

Plot[{f[x],F[x]),{x, a, b}] 

x/.Map[FindRoot[F'[x]==0, {x, #}] &,{2, 3, 5, 6}] 

x/.Map[FindRoot[f [x]==0, {x, #}] &,{ 1, 2, 4, 5, 6}] 
Slightly alter above commands for 75 - 80. 

Clear[x, f, F, u] 

a=0; f[x_] = x 2 - 2x - 3 

u[x_] = 1 — X 2 

F[x_] = Integrate [f[t], {t, a, u(x)}] 

x/.Map[FindRoot[F'[x]==0,{x,#}]&,{l,2,3,4)] 

x/.Map[FindRoot[F"[x]==0,{x,#}] &,{ 1, 2, 3, 4}] 
After determining an appropriate value for b, the following can be entered 

b = 4; 

Plot[{F[x],{x,a,b}] 

5.5 INDEFINITE INTEGRALS AND THE SUBSTITUTION RULE 

1 . Let u = 3x => du = 3 dx => | du = dx 

I sin 3x dx = I | sin u du = — | cos u + C = — | cos 3x + C 



2. Let u = 2x 2 =>- du = 4x dx =>■ \ du = x dx 



I x sin (2x 2 ) dx = / I sin u 



du = — | cos u + C = — j cos 2x^ + C 



3. Let u = 2t =^ du = 2 dt => \ du = dt 

I sec 2t tan 2t dt = I | sec u tan u du = | sec u 4- C = \ sec 2t + C 

4. Let u = 1 — cos ^ =>- du = \ sin | dt => 2 du = sin | dt 

J(l -cos |) 2 (sin i) dt= J 2u 2 du = § u 3 + C = § (l -cos |) 3 + C 

5. Let u = 7x — 2 =4> du = 7 dx => ^ du = dx 

J" 28(7x - 2y 5 dx = J" i (28)u~ 5 du = J" 4u~ 5 du = -u~ 4 + C = -(7x - 2)~ 4 + C 



6. Let u = x 4 - 1 => du = 4x 3 dx =^ \ du = x 3 dx 



/ x 3 (x 4 - l) 2 dx = J \ u 2 du = si + C = i (x 4 - l) 3 + C 



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324 Chapter 5 Integration 

7. Let u = 1 - r 3 =>• du = -3r 2 dr => -3 du = 9r 2 dr 

f 9fd^ = r _ 3u -i/2 du = _ 3 (2) u i/2 + C =-6(1- r 3 ) 1/2 + C 

8. Let u = y 4 + 4y 2 + 1 => du = (4y 3 + 8y) dy => 3 du = 12 (y 3 + 2y) dy 

J" 12 (y 4 + 4y 2 + l) 2 (y 3 + 2y) dy = J 3u 2 du = u 3 + C = (y 4 + 4y 2 + l) 3 + C 

9. Let u = x 3 / 2 - 1 =>■ du = § x 1 / 2 dx =>■ | du = y/x dx 

J ,/x sin 2 (x 3 / 2 - 1) dx = J | sin 2 u du = | (§ - i sin 2u) + C = i (x 3 / 2 - l) - 1 sin (2x 3 / 2 - 2) + C 



10. Let u = -i =4> du = - 2 dx 

I ^ cos 2 (i) dx = / cos 2 (— u) du = I cos 2 (u) du 



J + isin2u)+C = -^ + isin(-; 



£-*•*(!) +c 



C = - 1 cot 2 26> + C 



11. (a) Let u = cot 261 =>■ du = -2 esc 2 20 d0 => - 1 du = esc 2 20 d0 

J esc 2 20 cot 20 d0 = - / \ u du = - \ (| ) + C = - f 
(b) Let u = esc 20 =>• du = -2 esc 20 cot 20 d0 => - \ du = esc 20 cot 20 d0 

J esc 2 20 cot 20 d0 = J - 1 u du = - \ (| ) + C = - f + C = - \ esc 2 20 + C 

12. (a) Let u = 5x + 8 => du = 5 dx =>■ | du = dx 

/7fe5 = /K7u) du =i/ u " /2du =H2u 1/2 )+C=lu 4 / 2 + C=| v ^T8 + C 
(b) Let u = v/5x + 8 =4> du = ± (5x + 8)" 1 / 2 (5) dx ^ § du 



/5x+8 



13. Let u = 3 - 2s => du = -2 ds =4> - \ du = ds 



/ ^3^ ds = / ^u" (- \ du) = - \ JV/ 2 du = (- \) (| u 3 / 2 ) + C = - \ (3 - 2s) 3 / 2 + C 



14. Let u = 2x + 1 => du = 2 dx =$■ \ du = dx 



/ (2x + l) 3 dx = / u 3 (i du) = \ J u 3 du = (I) (£) + C = | (2x + l) 4 + C 



15. Let u = 5s + 4 => du = 5 ds =>• ± du = ds 

/^ds = /^(Idu)=i/u-V 2 du=(i)(2uV 2 ) +C =|y^T^ + C 

16. Let u = 2 — x =>- du= — dx => — du = dx 

J(2^dx = /^ = -3/u- 2 du=-3(^) + C=^ + C 

17. Let u = 1 - 2 =4> du = -20 d0 => - 1 du = d0 

/ VT 3 ^ 2 d0 = / \/u" (- 1 du) = - I / u 1 / 4 du = (- I) (I u 5 / 4 ) + C = - § (1 - 2 ) 5/4 + C 

18. Let u = 2 - 1 => du = 20 d0 => 4 du = 80 d0 

/ 80 Vfl 2 - 1 66 = f Vu"(4 du) = 4 / u 1 / 3 du = 4 (| u 4 / 3 ) + C = 3 (0 2 - 1) 4/3 + C 



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Section 5.5 Indefinite Integrals and the Substitution Rule 325 

19. Let u = 7 - 3y 2 => du = -6y dy => - \ du = 3y dy 

/3 y y7^3?dy = /^(-idu) =-i/u 1 /' 2 du= (-i) (f u 3 / 2 ) + C = - \ (7 - 3y 2 ) 3/2 + C 



20. Let u = 2y 2 + 1 => du = 4y dy 



/ 



4y dy 
V^y 2 + 1 



J 4- du = J u- 1 / 2 du = 2U 1 / 2 + C = 2v/2y 2 + 1 + C 



21. Let u = 1 + Jx =>■ du = 4r dx => 2 du = 4" 

v ^v x v x 



dx 



r i d X = r ^ 



22. Let u = 1 



i+\A 



+ C 



du 



J^^dx = /u 3 (2du) = 2(iu 4 )+C=i(l 



-V dx => 2 du = -4- dx 



23. Letu = 3z + 4 =>• du = 3 dz => 5 du = dz 



J cos (3z + 4) dz = I (cos u) (I du) = 5 J cos u du = | sin u + C = | sin (3z + 4) + C 



24. Let u = 8z - 5 =4> du = 8 dz => | du = dz 



I sin (8z — 5) dz = I (sin u) ( I du) = | I sin u du = | (—cos u) + C = — | cos (8z — 5) + C 

25. Let u = 3x + 2 =4> du = 3 dx => ± du = dx 

J sec 2 (3x + 2) dx = J (sec 2 u) (5 du) = \ J sec 2 u du = | tan u + C = | tan (3x + 2) + C 

26. Let u = tan x =>■ du = sec 2 x dx 

J tan 2 x sec 2 x dx = J u 2 du = | u 3 + C = 5 tan 3 x + C 

27. Letu = sin (|) =4> du = | cos (|) dx =>- 3 du = cos (|) dx 

f sin 5 (f ) cos (I) dx = / u 5 (3 du) = 3 (± u°) + C = \ sin G (f ) + C 

28. Let u = tan (|) => du = \ sec 2 (|) dx =^ 2 du = sec 2 (|) dx 
/tan 7 (I) sec 2 (I) dx = J u 7 (2 du) = 2 (± u 8 ) +C= 5 tan 8 (f) +C 

29. Let u = f| - 1 =^ du = | dr =>• 6 du = r 2 dr 

J r 2( f i„ 1 ) 5 dr= J u 5 (6du) = 6 J u 5 du = 6 ^) +C= ^_ 1 ) 6 +C 

30. Let u = 7 - ^ => du = - \ r 4 dr => -2 du = r 4 dr 
J r 4( 7 _ { i) 3 dr= J u 3 ( _2du) = -2/u 3 du=-2(^)+C=-l(7-ii) 4 + C 

31. Let u = x 3 / 2 + 1 => du = | x 1 / 2 dx =>■ | du = x 1 / 2 dx 

J x 1 / 2 sin (x 3 / 2 + 1) dx = J (sin u) (| du) = § J sin u du = § (-cos u) + C = - \ cos (x 3 / 2 + l) + C 



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326 Chapter 5 Integration 



du = | x 1 / 3 dx 



32. Let u = x 4 / 3 - 8 

J x 1 / 3 sin (x 4 / 3 - 8) dx = J (sin u) 



| du = x 1 / 3 dx 



du 



/ sin u du =| (—cos u) + C 



cos X 



4/3 



33. Let u = sec (v + |) => du = sec (v + f ) tan (v + f ) dv 

J sec (v + |) tan (v + |) dv = J du = u + C = sec (v + |) + C 

34. Letu = csc(^) => du = -icsc(^) cot(^) dv => -2du = csc(^) cot(^) dv 



/ 



esc (^j cot (^ 



*■) dv = J -2 du = -2u + C = -2 esc i 



^)+C 



35. Let u = cos (2t +1) =^ du = -2 sin (2t + 1) dt => - | du = sin (2t + 1) dt 



f ^ffttn * = f - i ^ = j- + C 

J cos 2 (2t+l) J 2 u 2 2u 



1 



2cos(2t+ 1) 



c 



36. Let u = 2 + sin t => du = cos t dt 



/ 



6 cos t 

(2 + sin t) = 



dt : 



/ | du = 6 / u~ 3 du = 6 (^|) + C = -3(2 + sin t)~ 2 + C 



37. Letu = coty =>- du = — esc 2 y dy =>- — du = esc 2 y dy 

J v/coTy esc 2 y dy = J v /u(-du) = - j u 1 / 2 du = - § u 3 / 2 + C = - f (cot y) 3 / 2 + C 



|(cot 3 y) 1/2 + C 



38. Let u = sec z => du = sec z tan z dz 



/ 



dz 



J i du = / u- 1 / 2 du = 2u J / 2 + C = 2^^^ + C 



39. Letu 



1 = t _1 - 1 =^ du = -t~ 2 dt => -du = i dt 



/ p cos (i — l) dt = I (cos u)(— du) = — I cos u du = —sin u + C = —sin Q — l) + C 



40. Let u = + 3 = t 1 / 2 + 3 => du = 1 1" 1 / 2 dt => 2 du = X dt 

J 4j cos (i/t + 3) dt = J (cos u)(2 du) = 2 J cos u du = 2 sin u + C = 2 sin (i/t + 3) + C 

41. Let u = sin \ => du = (cos j) (- gj) d# =4> -du = p cos | d# 
J p sin i cos | d6» = J -u du = - | u 2 + C = - i sin 2 | + C 

42. Let u = esc \f~6 => du = [-esc \/~6 cot y/Fj (~Krj d6* =>• -2 du = -4- cot \f~6 esc y^ d6> 

f /T wtt d <? = f 4s cot y/d esc y/9d6= f -2 du = -2u + C = -2 esc v^ + C = 

J Vf? sin 2 \/0 J vtf J sin 



43. Let u = s 3 + 2s 2 - 5s + 5 => du = (3s 2 + 4s - 5) ds 

JV + 2s 2 - 5s + 5) (3s 2 + 4s - 5) ds = / u du = f + C = ( s3 + 2s2 - 5s + 5 > 2 + C 

44. Let u = d i - 26 2 + W - 2 => du = (46> 3 - 46 + 8) dd =>• 1 du = (6» 3 - 6> + 2) d<9 

/ (0 4 - 20 2 + 89 - 2) (0 3 - 6 + 2) d# = / u (| du) = \ J u du = \ (f ) + C = ( g4 - 2fl2 + 8g - 2 ) 2 + C 



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Section 5.5 Indefinite Integrals and the Substitution Rule 327 



45. Let u = 1 + t 4 =>• du = 4t 3 dt =^ \ du = t 3 dt 



Jt 3 (l+t 4 ) 3 dt=/u 3 (|du) 



HX) 



C=i(l+t 4 ) 4 + C 



46. Let u = 1 - 1 => du = 4j dx 

^dx 



J# dx =J? 



/^JTTidx = /^du = /uV 2d u=|u3/ 2 + C=|(l-I) 3/2 + C 



47. Let u = x 2 + 1. Then du = 2xdx and |du = xdx and x 2 = u — 1. Thus I x 3 y x 2 + 1 dx = / (u — l)|-^/u 
= i/(u 3 / 2 - u 1 / 2 )du = i [|u 5 / 2 - |u 3 / 2 ] + C = |u 5 / 2 - |u 3 / 2 + C = |(x 2 + 1) 5/2 - i(x 2 + if' 2 



du 



+ C 



48. Letu = x 3 + 1 =^ du = 3x 2 dx and x 3 = u - 1. So f3x 5 y/x. 3 + ldx= J (u - l)y / udu = J(u 3/2 - u 1 / 2 ) 



du 



2 u 5/2 _ 2 u 3/2 + C = 2 (x 3 + ^5/2 _ 2,3 + ^3/2 + c 



(a) Let u = tan x => du = sec 2 x dx; v = u 3 =>• dv = 3u 2 du =>• 6 dv = 18u 2 du; w = 2 + v =>• dw = dv 



/ 



dx = f -^- 2 du = f ^^v = f 6*v = 6 f -2 d _ 6 -1 + C = _ ^_ + C 

J (2 + u 3 ) J (2 + v) 2 J w 2 J 2 + v 



18 tan 2 x sec 2 x 

(2 + tan 3 x) 2 " A ~~ J (2 + u 3 ) 2 "" ~ J (2 + v) 2 

= - 2+TP +C= — 2 + tan 3 x + ( ~' 

i3 v _v ^n 'J ton2 v e ~~2 v -J v — v. /: J.. lO *-or»2 „ „~~2 . 



(b) Let u = tan 3 x =>- du = 3 tan x sec x dx =>• 6 du = 18 tan x sec - x dx; v = 2 + u =>- dv = du 



J 1 



8 tan 2 x sec 2 x j v / 6 du l 6 dv 6 i /-i 6 i /-i 

QX — J (2 + u) 2 _ J v 2 ■ - + L ' "" +L 



(2 + tan 3 x) 2 J (2 + u) 2 J v 2 v ' *- 2 + u ' *- 2 + tan 3 x 

(c) Let u = 2 + tan 3 x =>■ du = 3 tan 2 x sec 2 x dx =>• 6 du = 18 tan 2 x sec 2 x dx 



+ C 



/ 1 8 tan 2 x sec 2 x j v / 6 du _ 6 ■ p 
(2 + tan 3 x) 2 QX ~ J u2 - u + ^ 



2 + tan 3 x 



50. (a) Let u = x — 1 => du = dx; v = sin u => dv = cos u du; w = 1 + v 2 =>• dw = 2v dv =>• | dw = v dv 
I i/l + sin 2 (x — 1) sin (x — 1) cos (x — 1) dx = I y 1 + sin 2 u sin u cos u du = J vy 1 + v 2 dv 

= J" 1 yV dw = 1 w 3 / 2 + C = i (1 + v 2 ) 3/2 + C = i (1 + sin 2 u) 3/2 + C = i (1 + sin 2 (x - 1)) 3/2 + C 

(b) Let u = sin (x — 1) =>• du = cos (x — 1) dx; v = 1 + u 2 => dv = 2u du =>• | dv = u du 

J a/1 +sin 2 (x - 1) sin (x - 1) cos (x - 1) dx = J u yTTu 2 du=Jiy / vdv = Ji v 1 / 2 dv 

= (I (I) v3/2 ) + C = 5 v3/2 + C = 5 (1 + u 2 ) 3/2 + C = i (1 + sin 2 (x - 1)) 3/2 + C 

(c) Let u = 1 + sin 2 (x — 1) =>• du = 2 sin (x — 1) cos (x — 1) dx =>• \ du = sin (x — 1) cos (x — 1) dx 

f y/l + sin 2 (x- l)sin(x- l)cos(x- 1) dx = J±y / u~du = J \ u 1 / 2 du = \ (|u 3 / 2 ) +C 
= i(l + sin 2 (x- l)) 3/2 + C 



51. Let u = 3(2r - l) 2 + 6 =>• du = 6(2r - 1)(2) dr => ^ du = (2r - 1) dr; v = y 7 ! 7 ^> dv = ^ du => ± dv 
du 



/ 



l 



(2r- l)cos V3(2r-l) 2 + 6 



dr 



/ 



co^/uX / i 



v / 3(2r-l) 2 + 6 "* J V V 7 " ^ ' '" 

i sin v/3(2r- l) 2 + 6 + C 



du 



J (cos v) (i dv) = 1 sin v + C = \ sin y 7 ^ + C 



52. Let u = cos y/~d => du = (-sin y^) f ^O d6» =>■ -2 du = 5^ d# 



f , sin ^ M= \ si "^ d# = f ^?# = -2 f u- 3 / 2 du = -2 (-2u- 1 /2) + C = 4 + C 



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328 Chapter 5 Integration 



y cos v 8 



+ c 



53. Let u = 3t 2 - 1 => du = 6t dt =4> 2 du = 12t dt 

s = Jl2t(3t 2 - l) 3 dt= Ju 3 (2du) = 2(|u 4 ) +C= |u 4 + C= \ (3t 2 - 1) 4 + C; 

s = 3 when t=l => 3 = ± (3 - l) 4 + C =5> 3 = 8 + C =4> C = -5 => s = \ (3t 2 - l) 4 - 5 



54. Let u = x 2 



du = 2x dx =>• 2 du = 4x dx 



y = J 4x (x 2 + 8)~ 1/3 dx = J u- J /3 (2 du) = 2(1 u 2 / 3 ) + C = 3u 2 / 3 + C = 3 (x 2 



y = when x = => = 3(8) 2 / 3 +C => C = -12 =^> y = 3(x 2 



\ 2 /3 



12 



\ 2 / 3 



C; 



55. Letu = t 



I> 



du = dt 



/ 8 sin 2 u du = 8 (jj - 1 sin 2u) + C = 4 (t + f0 - 2 sin (2t + § ) + C; 



8 when t = 



2 sin I 



=> s = 4(t+f^) -2sin(2t 



+ C =4> C = 8 



4t-2sin(2t+ |) +9 



56. Let u = | - 9 => -du = d6> 

r = J 3 cos 2 (| - 0) &6 = - J" 3 cos 2 u du = -3 (| + \ sin 2u) + C 



cos 
when = 



3jr 
8 



c 



-§(f-0)-f sin (f-20)+C; 



r =-i(f-^) 



3 



3 



29) + § 



| cos 26> + f + 



-20) + 



57. Let u = 2t - S => du = 2 dt =>• -2 du = -4 dt 



2 cos (2t - f ) 



| = J -4 sin (2t - |) dt = J (sin u)(-2 du) = 2 cos u + Ci = 2 cos (2t - §) + Ci 
at t = and g = 100 we have 100 = 2 cos (- §) + Ci => Ci = 100 => s 

=4> s = J* (2 cos (2t - |) + 100) dt = J (cos u + 50) du = sin u + 50u + C 2 = sin (2t - § 
at t = and s = we have = sin (- §) + 50 (- f ) + C 2 => C 2 = 1 + 25?r 

=>• s = sin (2t - f ) + 100t - 25tt + (1 + 25tt) => s = sin (2t - § ) + lOOt + 1 



100 



50(2t-|)+C 2 ; 



58. Let u = tan 2x => du = 2 sec 2 2x dx =>- 2 du = 4 sec 2 2x dx; v = 2x => dv = 2 dx => h dv = dx 



!| = J* 4 sec 2 2x tan 2x dx = J" u(2 du) = u 2 + d = tan 2 2x + C x ; 

dy 

dx 



at x = and ^ = 4 we have 4 = + Ci => Ci = 4 => f£ = tan 2 2x + 4 = (sec 2 2x - 1) + 4 = sec 2 2x + 3 
^ y = J (sec 2 2x + 3) dx = J (sec 2 v + 3) (± dv) = \ tan v + § 
at x = and y = - 1 we have - 1 = ± (0) + + C 2 =>■ C 2 = - 1 =>• y 



C 2 = 5 tan2x+3x + C 2 ; 



i tan 2x + 3x - 1 



59. Let u = 2t =» du = 2 dt =4> 3 du = 6 dt 

s = J 6 sin 2t dt = J (sin u)(3 du) = -3 cos u + C = -3 cos 2t + C; 

at t = and s = we have = -3 cos + C => C = 3 =4> s = 3 - 3 cos 2t 



s (|) =3 — 3 cos(7r) = 6 m 



60. Let u = 7rt => du = 7r dt => tt du = 7r 2 dt 

v = I 7r 2 cos 7rt dt = I (cos u)(7r du) = tt sin u + Ci = 7r sin (7rt) + Ci ; 
at t = and v = 8 we have 8 = 7r(0) + Ci =4> Ci = 8 => v = ^ = n sin (7rt) 
= J sin u du + 8t + C 2 = -cos (7rt) + 8t + C 2 ; at t = and s = we have = 



! => s = J (tt sin (7rt) + 8) dt 
•1+Ca =>■ C 2 = 1 



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Section 5.6 Substitution and Area Between Curves 329 



S = 8t - COS (7Tt) +1 => s(l) = 8 - COS 7T + 1 = 10 m 



61. All three integrations are correct. In each case, the derivative of the function on the right is the integrand on 
the left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover, 



sin 2 x + Ci = 1 - cos 2 x + Ci => C 2 = 1 + Ci ; also -cos 2 x + C 2 



,2.x 



c 2 



Ci 



62. Both integrations are correct. In each case, the derivative of the function on the right is the integrand on the 
left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover, 



tajp + C 



sec x— 1 



/-i sec x 



a constant 



63. (a) (xTq) J V max sin 12U7rt dt = 60 [-V„ (j^) cos (1207rt)] J 760 = - ¥=■ [cos 2tt - cos 0] 



= -^[l-l]=0 

(b) V max = \[2N m = ^(240) w 339 volts 

(c) J (V max ) 2 sin 2 1207rtdt = (V max ) 2 J 

1/60 _ (V,,,,,) 2 



1 - cos 240-tt t ■> 



dt 



(v- 



- ,.1/60 

fL J g ( 



(1 - cos 2407rt) dt 



t 



Ilk) sin240^t Jo 



[(§13 ~ (21k) sin ( 4 ^)) - (0- (21k) s in (0))] = ^ 



20 



5.6 SUBSTITUTION AND AREA BETWEEN CURVES 



1. (a) Letu = y+1 => du = dy; y = => u = 1, y = 3 =4> u = 4 

J 3 VV+i d y = T ul/2 du = [f u3/2 ] 1 = (I) (4)3/2 ~ (I) (1)3/2 



(8) 



:u<i)=£ 



(b) Use the same substitution for u as in part (a); y = — 1 =>• u = 0, y = => u = 1 



£y>7TTdy = X 1 uV 2 du=[lu3/ 2 ];=(|)(l)3/ 2 -0 



2. (a) Let u = 1 - r 2 => du = -2r dr => - 5 du = r dr; r = =>• u=l,r=l =>• u = 

£rv / T^ 2 dr= ^°-i v A 1 du= [- | u 3 / 2 ] J = - (- ±) (l) 3 / 2 = § 
(b) Use the same substitution for u as in part (a); r = — 1 =>• u = 0, r = 1 => u = 

J_ 1 i r/r^dr=X°-i v AIdu = 



3. (a) Let u = tan x =>• du = sec 2 x dx; x = => u = 0, x = f =>• u=l 



/W 4 
I t 

Jo 



tan x sec - x dx 



Jo 



du 



S-o 



(b) Use the same substitution as in part (a); x 



■ l,x = =>• u = 



I" 

<J — 7rA 



tan x sec x dx 



/> 



du 



4. (a) Let u = cos x => du = - 
I 3 cos 2 x sin x dx = I 



sin x dx => — du = sin x dx; x = => u = 1, x = 7r => u 

-1 
-3u 2 du = f-u 3 l 7 1 



(b) Use the same substitution as in part (a); x = 2tt 

-3u 2 du = 2 



_(_1)8 _ (_ ( 1)8) = 2 

U=1,X=37T =>■ U = — 1 



J 3 cos 2 x sin x dx = I 
2-k J 1 



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330 Chapter 5 Integration 



5. (a) u = 1 + t 4 => du = 4t 3 dt => \ du = t 3 dt; t = => u = 1, t = 1 =>• u = 2 



21 _ il _ I! 

16 16 16 



(b) Use the same substitution as in part (a); t = — 1 =>■ u = 2, t = 1 =>• u = 2 
J t 3 (1 + t 4 ) 3 dt = j \ u 3 du = 



=>■ du = 2t dt =^> i du = t dt; t = => u = 1, t = \fl 

•J u «/ 1 

(b) Use the same substitution 



6. (a) Let u = t 2 + 1 

J (i 



= t 2 + 1 =4> du = 2t dt =4> i du = t dt; t = =4> u = 1, t = a/7 => u = 8 

t { e + 1) 1/3 dt = £ \ U V3 du = [(i) (|) u 4 / 3 ] ; = (i) ( 8) 4 / 3 - (|) (1) 4 / 3 = f 

he same substitution as in part (a); t = — y 7 => u = 8, t = =>■ u=l 
/ 3 du=-J^ 8 iu 1/3du =-f 



._..,■. .nv. ,.„,„. ....v. ...,,v..,v, u as in part (a); t = -\/7 => u = 8, t = =>■ u=l 

v< t2+1) dt =J> 1/0 " "' " 

7. (a) Let u = 4 + r 2 



du = 2r dr =► ^ du = r dr; r = — 1 =>• u = 5, r = 1 =>• u = 5 
/ — ^ 



dr 



= 5 XV 



du = 



(b) Use the same substitution as in part (a); r = => u = 4, r = 1 => u = 5 

r^ dr = 5 r^- 2 du = 5[-iu-^ = 5(-i(5r 4 )-5(-i(4ri)=I 

(a) Let u = 1 + v 3 / 2 => du = | v 1 / 2 dv =>■ ^ du = lOy^v dv;v = => u=l,v=l => u 

p 1 r- n 2 n 2 

I lOv/v , ( i n(\ , \ ■■il i _■■ , in r : - ■ - >ii i i n in 

Jo ft- 



,3 /2f dV=j; 2 ^(fdu)=f / 2 U-MU: :- ^7 :■■ ^ "~™ 



3 LuJ 1 ~ 3 L2 1 J 3 



(b) Use the same substitution as in part (a); v=l => u = 2, v = 4 =>• u = 1 + 4 3 / 2 = 9 

du ) = _ 20 r 11 9 _ 20 f 1 



P !°V^ dv _ f 9 J_ f20 

j\ (l +v 3/2) 2 QV ~ J 2 " 2 V3 

9. (a) Let u = x 2 + 1 -r uu-^u* -^ 

I ^=dx= J 4-du = 2U- 1 



3 LuJ 2 



20 fl _ 1\ _ _ 20 /_ J7_\ _ 70 
3 V9 2^ — 3 I 18) ~ 27 



u = 1, x = \/3 =4> u = 



du = 2x dx =4- 2 du = 4x dx; x = 

rV2 du = [4u 4 / 2 ] J = 4(4) J / 2 - 4(l) 1 /2 = 
(b) Use the same substitution ; 



as in 

/v/3 n4 

-v/3 \/x 2 + l J 4 v/ U 



10. (a) Let u = x 4 + 9 



part (a); x = — \/ 3 => u = 4, x = \/3 =>- u 







=> du = 4x 3 dx => i du = x 3 dx; x = =>■ u = 9, x = 1 =>• u = 10 

/.' 7^ dx = X' 1 "- 1/2 du = [I ( 2 ) ul/2 ] J° = I d0) 1/2 - 2- (9) 1/2 = #=2 



■., v/x'i+9 "'" J9 4" — L4V- J 9 2-"' 2^> 2 

(b) Use the same substitution as in part (a); x = — 1 =>■ u = 10, x = =>• u = 9 

^^dx= / i U - 1 / 2 du=- I U - 1 /2 d U= 3 ^^ 

-l \/x4 + 9 J 10 4 J 9 4 2 



11. (a) Let u = 1 - cos 3t =^> du = 3 sin 3t dt ^ \ du = sin 3t dt; t = => u = 0, t = § =^> u = 1 - cos | = 1 

£ /6 (l-cos3t)sin3tdt=/ o 1 iudu=[i(f)]^=i(l) 2 -I(0) 2 =i 

(b) Use the same substitution as in part (a); t = | =£> u=l,t=| => u=l — cos 7r = 2 
■ v ft 

(1 — cos 3t) sin 3t ■ 



f-Tft 

Jx/6 



u 

^t=/ju^=[i(f)]; : 



1(2)^-1(1)2 = i 



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Section 5.6 Substitution and Area Between Curves 33 1 



12. (a) Let u = 2 + tan \ => du = \ sec 2 | dt => 2 du = sec 2 | dt; t = =^ => u = 2 + tan (=^) = 1, t = =$> u = 2 

J ^ (2 + tan ±) sec 2 \ dt = J u (2 du) = [u 2 ] J = 2 2 - l 2 = 3 



(b) Use the same substitution as in part (a); t = -£ => u=l,t=| => u = 3 
J"" 2 (2 + tan I) sec 2 \ dt = 2 J u du = [u 2 ]j = 3 2 l 2 = 8 



13. (a) Let u = 4 + 3 sin z =4> du = 3 cos z dz =>• J du = cos zdz;z = =>• u = 4, z = 2n =4> u = 4 



/A cos *. dz = / 4- (A du) = 

Jo y 4 + 3 sin z J 4 V" v ' ' 



(b) Use the same substitution as in part (a); z = — 7r =4> u = 4 + 3 sin (— 7r) = 4, z = ir =>■ u = 4 



£ 



x/4 + 3! 



dz 



r^(5 du )=° 



14. (a) Let u = 3 + 2 cos w =$■ du = —2 sin w dw =>• — | du = sin w dw; w = — f =>■ u = 3, w = =>• u = 5 






dw 



r» 



--' 1 du) = ifu- 1 i 5 - ±fl n 



15 



_^2 (3 + 2cosw) 2 " vv J 3 " V 2 ""7 2 L" J3 2 V5 3 V 

(b) Use the same substitution as in part (a); w = =>■ u = 5, w = | =>■ u = 3 
r „J nw a dw = f V 2 (- | du) = i f u- 2 du = i 

Jo (3 + 2 cos w) 2 J 5 V 2 / 2 J 15 



15. Let u = t 5 + 2t =>■ du = (5t 4 + 2) dt; t = =>■ u = 0, t = 1 =4> u = 3 

/ o ' y/W+K (5t 4 + 2) dt = JV/ 2 do = [| u 3 / 2 ] J = f (3) 3 / 2 - f (0) 3 / 2 =2^ 



16. Letu=l+ v /y =4> du=^;y = l => u = 2, y = 4 =>• u 



f 4 ' y ^ = f 3 \du= fV 2 du=[-u- 1 ]2 

J i 2J^(l + Jyi 2 J2 u2 ^ L J2 



2^ 6 



17. Let u = cos 261 => du = -2 sin 20 d0 =>• - § du = sin 20 d0; = => u = 1, = | => u = cos 2 (f ) = \ 

r^ 6 r 1/2 r 1/2 r / _,m i/ 2 

J o cos- 3 20 sin 20 d0 = J L u - 3 (-idu) = -lJ | u~ 3 du = [- \ (^)] j 



4 (i) 2 4(1)2 4 



18. Let u = tan (I) => 



u = tan | = 1 



du = 1 sec 2 (|) d0 => 6 du = sec 2 (§) d0; = tt =>■ u = tan (f ) = -A- . 



ni*/2 



^t 5 (f) sec 2 (f ; 



« w =C u " 5(6du) =[ 6 (^)]U=[-* ] 



2u<U 1/^/3 



3 
2(1)4 



12 



.s/i) 



19. Let u = 5 - 4 cos t =4> du = 4 sin t dt =>■ | du = sin t dt; t = =^ u = 5 - 4 cos = 1, t = n =4> 
u = 5 — 4 cos 7r = 9 

J*5 (5-4 cos t) 1 / 4 sin t dt = J^u 1 / 4 (A du) = | J^u 1 / 4 du = [§ (f u 5 / 4 )] J = 9 5 / 4 - 1 = 3 5 / 2 - 1 

20. Let u = 1 - sin 2t => du = -2 cos 2t dt =>■ - \ du = cos 2t dt; t = =4> u = 1, t = | => u = 
/^(l - sin 2t) 3 / 2 cos 2t dt = f- 1 u 3 / 2 du = [- A (§ u 5 / 2 )] J = (- 1 (Of' 2 ) - (- A (l) 5 / 2 ) = \ 



21. Letu = 4y-y 2 + 4y 3 + 1 => du = (4 - 2y + 12y 2 ) dy; y = =4> u = 1, y = 1 => u = 4(1) - (l) 2 + 4(1) 3 + 1 
J (4y - y 2 + 4 y 3 + 1)- 2 / 3 ( 12y 2 _ 2y + 4) dy = J] u- 2 / 3 du = [3u 4 / 3 ] \ = 3(8) 4 / 3 - 3(1) 4 / 3 = 3 

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332 Chapter 5 Integration 

22. Let u = y 3 + 6y 2 - 12y + 9 =4> du = (3y 2 + 12y - 12) dy =>• | du = (y 2 + 4y - 4) dy; y = => u = 9, y = 1 
=> u = 4 

J o '(y 3 + 6y 2 - 12y + 9)" 1/2 (y 2 + 4y - 4) dy = f f \ u- 1 ' 2 du = [I ^u 1 ^)] J = | (4)V2 _ | (9) i/2 = | ( 2 _ 3) 

_ 2 
3 



23. Let u = 6» 3 / 2 =>- du = 1 (9 1 / 2 d0 =s> | du = \/9 d0; 9 = => u = 0, 6» = Vtt 2 ^ u = tt 



J* V "v/0 cos 2 (# 3 / 2 ) dfl = JJcos 2 u (I du) = [ 



2 fu , I 

3 l2 T 4 



2u)] 



~~ 3 \2 "^ 4 



sin 2tt) - I (0) = f 



3 v"; — 3 



24. Let u = 1 + \ =^ du = -t~ 2 dt; t = — 1 => u = 0, t = - \ => u = -1 



£~'V 2 sin 2 (1 + I) dt = Jf-sin 2 udu =[-(!- 1 sin 2u)] 



- U - 5 - | sin 



'(-2)) 



'0 _ 1 
v2 4 



sin 



0) 



\ - \ sin 2 



25. Let u = 4 - x 2 =4> du = -2x dx => - | du = x dx; x = -2 => u = 0, x = =>■ u = 4, x = 2 => u = 

A = - J x^4 - x 2 dx + J x^4 - x 2 dx = - J - ± u 1 / 2 du + J - \ u 1 / 2 du = 2 J \ u 1 / 2 du = J u 1 / 2 du 

= [I u3/2 ] o = I W 3/2 " I <°) 3/2 = f 



26. Let u = 1 — cos x =4> du = sin x dx; x = =>- u = 0, x = 7r =>• u = 2 

2 



f 



f 1 - cos x) sin x dx = J u du = | 4 I = =f - -, : 2 



(i 



2 2 



27. Let u = 1 + cos x =>■ du = —sin x dx => — du = sin x dx; x = —ir =^ u = 1 + cos (— ir) = 0, x = 

=>■ u = 1 + cos = 2 

A = - J_ 3 (sin x) a/1 +cosx dx = - J 3U 1 / 2 (-du) = 3 J u 1 / 2 du = [2u 3 / 2 ] 2 = 2(2) 3 / 2 - 2(0) 3 / 2 = 2 5 / 2 

28. Let u = 7r + 7r sin x =>• du = 7r cos x dx =>■ ^ du = cos x dx; x = — | =>■ u = 7r + 7r sin (— f ) = 0, x = =>• u = 7r 

| (cos x) (sin (7T + 7r sin x)) dx = 2 I | (sin u) (- du) 

= I sin u du = [—cos u]q = (—cos jf) — (—cos 0) = 2 



1 — cos 2x . 



29. For the sketch given, a = 0, b = n; f(x) — g(x) = 1 — cos x = sin x — 

A= J**iL = »!M dx= I < £ ,r (1 _ COS 2 X )dx=i[x-^];=i[(7r-0)-(0-0)]=| 



30. For the sketch given, a 



2f 



sec 2 t + 4 sin 2 t dt 



/lr/3 
-ir/ 3 

I sec 2 1 dt + 2 | 

*/ — 7r/3 «./ — rr 



f ; f(t) - g(t) = \ sec 2 1 - (-4 sin 2 1) = \ sec 2 1 + 4 sin 2 t 

/tt/3 /-»7r/3 P 71 "/^ p 71 "/^ 

sec 2 1 dt + 4 sin 2 1 dt = \\ sec 2 1 dt + 4 (1 " c o 0s2t) dt 
-tt/3 J-jr/3 2 J-tt/1 J -tt/: 



Kfi 



■k/3 



,, , ^sec 2 1 dt + 2j_ n (l - cos 2t) dt = \ [tan t]_J /3 + 2[t - ^j"'" = v /3+4-f- v / 3=f 



31. For the sketch given, a = -2, b = 2; f(x) - g(x) = 2x 2 - (x 4 - 2x 2 ) = 4x 2 - x 4 ; 



A = f ^ (4x 2 - x 4 ) dx = [ 



4x 3 _ x^ 
3 5 



'32 _ 32 \ 



32 
3 



32 M _ 64 64 _ 320-192 _ 128 



! )] 



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Section 5.6 Substitution and Area Between Curves 333 



32. For the sketch given, c = 0, d = 1; f(y) — g(y) = y 2 — y 3 ; 



(1-0) _ (1-0) _ 1 _ 1 _ J_ 
3 4 — 3 4 — 12 



33. For the sketch given, c = 0, d = 1; f(y) - g(y) = (12y 2 - 12y 3 ) - (2y 2 - 2y) = 10y 2 - 12y 3 + 2y; 

A=/ o '(10y 2 -12 y 3 + 2 y )d y = / o I 10 y 2 dy-/ o 1 12 y 3 dy + / o '2ydy=[fy3];~[fy4]; + [|y 2 ]; 
= (f -0)- (3-0) + (1-0) = | 



34. For the sketch given, a = -1, b = 1; f(x) - g(x) = x 2 - (-2x 4 ) = x 2 + 2x 4 ; 



A 



J> + 2x 4 )dx=[f + f]^ = (i + f) -[-i + (- 2 )] 



3^5 



10+12 _ 22 
15 ~ 15 



35. We want the area between the line y = 1, < x < 2, and the curve y = \, minus the area of a triangle 



(formed by y = x and y = 1) with base 1 and height 1. Thus, A 



I'M) 



dx-i(l)(l) 



= (2- J M-i=2- 2 -- i = 5 

K* 12 J 2 A 3 2 6 

36. We want the area between the x-axis and the curve y = x 2 , < x < 1 plus the area of a triangle (formed by x = 1, 



x + y = 2, and the x-axis) with base 1 and height 1 . Thus, A 



Jo 



dx+i(l)(l) 



1 _ 1 , 1 _ 5 

2 ~~ 3 "r 2 ~~ 6 



37. AREA = Al + A2 

Al: For the sketch given, a = — 3 and we find b by solving the equations y = x 2 — 4 and y 



2x 



simultaneously for x: x 2 - 4 = -x 2 - 2x =4> 2x 2 + 2x - 4 = => 2(x + 2)(x - 1) => x = -2 or x = 1 so 



b = -2: f(x) - g(x) = (x 2 - 4) - (-x 2 - 2x) = 2x 2 + 2x - 4 => Al 



/> 



x 2 + 2x - 4) dx 



¥ + ¥-4*i v , 



16 ' 4+8) -(-18 + 9+ 12) = 9- 16 - " 



A2: For the sketch given, a = -2 and b = 1: f(x) - g(x) = (-x 2 - 2x) - (x 2 - 4) = -2x 2 - 2x + 4 



3 — 3 
2 _ o v \ _ f v 2 _ A\ _ _o v 2 



=> A2 = - 



J_ n (2x 2 + 2x - 4) dx = - Uf + x 2 - 4x1 = - (§ + 1 - 4) 



;-f+ 4 - 



5-1+4-^+4 + 8 = 9; 



3S 



Therefore, AREA = Al+A2=-^+9- , 



38. AREA = Al + A2 

Al: For the sketch given, a = -2 and b = 0: f(x) - g(x) = (2x 3 - x 2 - 5x) - (-x 2 + 3x) = 2x 3 - 8x 



Al 



£(2x3 



8x) dx 



2x 4 8x 2 



0-(8- 16) = 8; 



A2: For the sketch given, a = and b = 2: f(x) - g(x) = (-x 2 + 3x) - (2x 3 - x 2 - 5x) = 8x - 2x 3 

=> A2 = j g (8x - 2x 3 ) dx = fe - ^1 = (16 - 8) = 8; 
Therefore, AREA = Al +A2 16 



39. AREA = Al + A2 + A3 

Al: For the sketch given, a = — 2 and b = — 1: f(x) — g(x) = (— x + 2) — (4 — x 2 ) = x 2 — x — 2 

=* Al = £ 1 (x 2 -x-2)dx=[f-f-2x]^=(-3-i+2)-(-f- 4 +4) 
A2: For the sketch given, a = - 1 and b = 2: f(x) - g(x) = (4 - x 2 ) - (-x + 2) = - (x 2 - x - 2) 



7 _ 1 _ 14-3 _ n. 
3 2 6 6 ' 



A2 



■/: < 



x 2 - x - 2) dx 



2x 



'8 _ 4 
v3 2 



4) 



l-i+2) 



-3 + 8- ^ - 2'' 



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334 Chapter 5 Integration 



A3: For the sketch given, a = 2 and b = 3: f(x) - g(x) = (-x + 2) - (4 - x 2 ) = x 2 - x - 2 



=>. A3 = J (x 2 - x - 2) dx 
Therefore, AREA = Al + A2 + A3 



2x 



6 



+ (9 



'27 



6) 



4) 



49 
6 



40. AREA = Al + A2 + A3 

Al: For the sketch given, a = -2 and b = 0: f(x) - g(x) = ( y - xj - § = y - | x = | (x 3 - 4x) 



Al 



£ (* 3 



4x) dx 



1 Ui 

3 4 



2x< 







-2 



3 (4- 8)= |; 



A2: For the sketch given, a = and we find b by solving the equations y 

x 3 „ x _ , x 3 4 



forx: j - x = | =>• y - j x = 
f(x)-g(x) = f-(f -x) =-|(x 



| (x - 2)(x + 2) = => x 
'' -4x) =>■ A2 = - i J (x 3 - 4x) dx = ± Jj (4x - x 3 ) = 1 |2x 



j — x and y = f simultaneously 
—2, x = 0, or x = 2 so b = 2: 

2 



4) 



A3: For the sketch given, a = 2 and b = 3: f(x) - g(x) = ( y - xj - | = | (x 3 - 4x) 

=► A3= 3 /;(x 3 -4x)dx=I^-2x 2 i;= 3 [( T -2.9)-(f-8)] 



I (f - 14) 



Therefore, AREA = Al + A2 + A3 



25 
12 



32+25 



19 

4 



25. 
12' 



41. a= -2, b = 2; 
f(x) - g(x) = 2 



(x 2 - 2) = 4 - x 2 



A= J_ 7 (4-x 2 )dx= Ux 



'24 



32 
3 




42. a= -1, b = 3; 

f(x) - g(x) = (2x - x 2 ) - (-3) 

=-. A = f (2x - x 2 + 3) dx = 
= (9-f + 9)-(l + i-3) 



2x 



11 



3x 

32 
3 



y=2x-x 2 




43. a = 0, b = 2; 

f(x) - g(x) = 8x 



2 



16 



32 
5 



= /> 

80-32 _ 48 
5 — 5 



x 4 )dx 




44. Limits of integration: x 2 — 2x = x =4> i 
=> x(x - 3) = => a = and b = 3; 
f(x) - g(x) = x - (x 2 - 2x) = 3x - x 2 



3x 



> A = 

¥-9 



J> 



x - x 2 ) dx = I ¥ - 4 



27-18 _ 9 
2 2 




Copyright (c) 2006 Pearson Education 




45. Limits of integration: x 2 = — x 2 + 4x =>• 2x 2 — 4x = 
=> 2x(x - 2) = =4> a = and b = 2; 
f(x) - g(x) = (-x 2 + 4x) - x 2 = -2x 2 + 4x 

J (-2x 2 + 4x) dx 

-32 + 48 _ 8 



Section 5.6 Substitution and Area Between Curves 

y 



335 



A 



-2x 3 | 4x^ 
3 + 2 



16 , 16 

3 + 2 




46. Limits of integration: 7 - 2x 2 = x 2 + 4 => 3x 2 - 3 = 
=>• 3(x- l)(x+ 1) = => a=-landb=l; 
f(x) - g(x) = (7 - 2x 2 ) - (x 2 + 4) = 3 - 3x 2 

=> A = J(3 - 3x 2 ) dx = 3 [x - f ] * 
= 3[(l-|)-(-H-I)]=6(l)=4 _1 



y=7-2x2 




47. Limits of integration: x 4 — 4x 2 + 4 = x 2 

=> x 4 - 5x 2 + 4 = => (x 2 - 4) (x 2 - 1) = 
=> (x + 2)(x - 2)(x + l)(x - 1) = => x = -2, -1, 1, 2; 
f(x) - g(x) = (x 4 - 4x 2 + 4) - x 2 = x 4 - 5x 2 + 4 and 
g(x) - f(x) = x 2 - (x 4 - 4x 2 + 4) = -x 4 + 5x 2 - 4 

=> A = J (-x 4 + 5x 2 - 4)dx + J (x 4 - 5x 2 + 4)dx 
+ J f (-x 4 + 5x 2 - 4)dx 



- :i 4x| ^ + [f - f + 4x] ^ + [=* + f - 4x] \ 




(i-| + 4)-(f-f + 8) + (I-f + 4)-(-I + f-4) + 

60 , 60 _ 300-180 _ o 

~ T + T - vT~ ~ 8 



32 , 40 
5 + 3 



1 + ^ 

5^3 



4) 



48. Limits of integration: xa/ a 2 — x 2 = =^ x = or 

\/a 2 - x 2 =0 =4> x = or a 2 - x 2 = =>■ x = -a, 0, a; 

A= / -xy 7 a 2 - x 2 dx + / xy 7 a 2 - x 2 



dx 



1 \2 /„2 „2n3/2 



, L ., ( a-- -xT 

i(a 2 r- -I(a^) 



I [| (a 2 - x 2 ) 3/i 




-x, x < 



x i 6 

5 T 5 



49. Limits of integration: y 

V x, x > 

5y = x + 6ory= | + f; for x < 0: \J — x 
=>■ 5^^ = x + 6 =^ 25(-x) = x 2 + 12x + 36 
=> x 2 + 37x + 36 = => (x + l)(x + 36) = 
=4> x = — 1, —36 (but x = —36 is not a solution); 

for x > 0: 5y/H = x + 6 => 25x = x 2 + 12x + 36 
=> x 2 - 13x + 36 = => (x - 4)(x - 9) = 
=> x = 4, 9; there are three intersection points and 

6 



and 



A 



j: 



x + 6 



^x+x^-v^+r^ 



y = V :: x 




4 s ) 



dx 



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336 Chapter 5 Integration 



(x + 6) 2 , 2 



10 



+ 1 (~*) 3/2 ] \ +[^f-l * 3/2 ] I + [f ^ " ^f] ' 



'36 _ 25 _ 2\ , / 100 _ 2 4 3/2 _ 36 , ft \ , (2 q 3/2 _ 225 _ 2 4 3/2 , K>0\ 
vlO 10 3) ' \ 10 3" H 10 1" U J ' I 3 ' * 10 3" H "I" 10 / 



50 , 20 5 

10 ' 3 3 



50. Limits of integration: 

x 2 - 4, x < -2 or x > 2 
4 - x 2 , -2 < x < 2 

for x < -2 and x > 2: x 2 - 4 = f + 4 

=> 2x 2 - 8 = x 2 + 8 = 
for -2 < x < 2: 4 - x 2 

=> x 2 = => x = 0; by symmetry of the graph, 




j/ = i 2 /2 + 4 



= 2(|-0)+2(32-f-16 + |) 



40-56-64 
tu 3 — 3 



51. Limits of integration: c = and d = 3; 

f(Y) - g(y) = 2y 2 - = 2y 2 



/ V dy 



2-9= 18 




1 — x 



52. Limits of integration: y 2 = y + 2 => (y + l)(y — 2) = 



- 1 and d 

-.2 



2; f(y) - g(y) = (y + 2) - y 2 



A 



J> 



+ 4- 



2 - y 2 ) dy = 



^+2y 
6-f- 




53. Limits of integration: 4x = y 2 — 4 and 4x = 16 + y 
=> y 2 -4=16 + y => y 2 -y-20 = => 
(y - 5)(y + 4) = => c = -4 and d = 5; 

f(y)-g(y)= C-T) - (^) = =^T^ 



80) 



A = 


= U_ 4 (-y 2 + y+20)d y 


if- 


^+20y] 5 4 


u- 

H- 


f + f + 100)-i(f + f 

f + 1 + 180) = ^ 




-y=16 



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54. Limits of integration: x = y 2 and x = 3 — 2y 2 

=> y 2 = 3 - 2y 2 => 3y 2 = 3 => 3(y - l)(y + 1) = 
=> c = -1 and d = 1; f(y) - g(y) = (3 - 2y 2 ) - y 2 

= 3 - 3y 2 = 3 (1 - y 2 ) => A = 3 fj\ - y 2 ) dy 



Section 5.6 Substitution and Area Between Curves 337 

y 



3-2(l-I)=4 



3(1-1) -3(-l + I) 



x+2y2=3 



x-y2=0 




55. Limits of integration: x = — y 2 and x = 2 — 3y 2 

=> -y 2 = 2 - 3y 2 =4> 2y 2 - 2 = 

=> 2(y - l)(y + 1) = => c = -1 and d = 1; 

f(Y) - g(Y) = (2 - 3y 2 ) - (-y 2 ) = 2 - 2y 2 = 2 (1 - y 2 ) 



=> A = 2/_ 1 | (l~y 2 )dy = 2[ 
= 2(l-I)-2(-l + i)=4 



y-i 



2\ _ 8 
3 ^ 3 



x+3y 2 =2 




56. Limits of integration: x = y 2 / 3 and x = 2 — y 4 

=> y 2 / 3 = 2 - y 4 => c = -1 and d = 1; 

f(y) - g(y) = (2 - y 4 ) - y 2/3 

=> A=/' i (2-y 4 -y 2 / 3 ) dy 

= [^y-i-h 5/3 }\ 

= (2-I-|)-(-2+I + |) 




2(2 



1 3\ _ 12 



57. Limits of integration: x = y 2 — 1 and x = |y| \/l — y 2 

=>. y 2 - 1 = |y| ^/1-y 2 => y 4 - 2y 2 + 1 = y 2 (1 - y 2 ) 

=^ y 4 - 2y 2 + 1 = y 2 - y 4 => 2y 4 - 3y 2 + 1 = 

=> (2y 2 - 1) (y 2 - 1) = => 2y 2 - 1 = or y 2 - 1 = 



v 2 = l „, „2 

y 2 



ory 2 =l => y= ± ^ or y = ± 1 . 



± /I 
Substitution shows that — 5j— are not solutions => y = ±1; 

for -1 < y < 0, f(x) - g(x) = -y y/l - y 2 - (y 2 - 1) 

= 1— y 2 — y (1 — y 2 ) , and by symmetry of the graph, 

A = 2£[l-y 2 -y(l-y 2 ) 1/2 ]dy 

= 2/_ i (l-y 2 )dy-2£° i y(l- y 2 ) 1/2 dy 



2y- 



'1\ |2(l-y 2 ) 3/2 



i 




x =1 y I \/i -y 2 



2[(0-0)-(-l + i)] + (|-0)=2 



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338 Chapter 5 Integration 



58. AREA = Al + A2 

Limits of integration: x = 2y and x = y 3 — y 2 => 

y 3 _ y 2 = 2 y =>. y (y 2 - y - 2) = y(y + l)(y - 2) = 
^ y = -1,0,2: 
for - 1 < y < 0, f(y) - g(y) = y 3 - y 2 - 2y 



=* Al = /_" i (y 3 -y 2 -2y)dy=[^-f 

= o-(i + J-i) = & 

for < y < 2, f(y) - g(y) = 2y - y 3 + y 2 

=> A2 = / o 2 (2y-" 3 ' •■-> 

Therefore, Al+A2=-^ + f = g 



r 



y J + y'J dy 



y 2 -^ 




59. Limits of integration: y = — 4x 2 + 4 and y = x 4 — 1 



1 = -4x 2 +4 => x 4 + 4x 2 







=> (x 2 + 5)(x- l)(x+ 1) = => a= -1 andb= 1; 
f(x) - g(x) = -4x 2 + 4 - x 4 + 1 = -4x 2 - x 4 + 5 



A 



Jj- 



4 _ 1 

3 5 



-4x 2 - 
5)- 



5)dx 



4x 3 
3 



+ 5x 



|-5)=2(-|-| + 5) 



-l 

104 
15 




60. Limits of integration: y = x 3 and y = 3x 2 — 4 

=^ x 3 - 3x 2 + 4 = => (x 2 - x - 2) (x - 2) = 
=> (x + l)(x - 2) 2 = => a = -1 and b = 2; 
f(x) - g(x) = x 3 - (3x 2 - 4) = x 3 - 3x 2 + 4 

2 

•4x1 



A = j (x 3 - 3x 2 + 4) dx = h 

)-a+i-4)=f 



3xf 

4 3 



16 _ 24 

4 3 




61. Limits of integration: x = 4 — 4y 2 and x = 1 — y 4 

=> 4 - 4y 2 = 1 - y 4 => y 4 - 4y 2 + 3 = 

=>• (y-v^) (y + v/3)(y-i)(y+i) = o =* c = -i 

and d = 1 since x > 0; f(y) - g(y) = (4 - 4y 2 ) - (1 - y 4 ) 
= 3 - 4y 2 + y 4 => A = J__(3 - 4y 2 + y 4 ) dy 



3y 



4y 3 



2(3-f + i) 



56 
15 



x =4-4y2 




x=1-y 4 



62. Limits of integration: x = 3 — y 2 and x 



y~ 



4 



3 = => | (y - 2)(y + 2) = 



-2 and d 



2;f(y)-g(y) = (3-y 2 )-(^) 

3(l-fl ^A = 3£(l-£)dy = 3[; 
3[(2-&)-(-2+£)]=3(4 



y-i 



\l) 



12-4 




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Section 5.6 Substitution and Area Between Curves 339 



63. a = 0, b = 7r; f(x) — g(x) = 2 sin x — sin 2x 

=> A = J (2 sin x - sin 2x) dx = [-2 cos x + ^] £ 
= [-2(-l)+i]-(-2-l + i)=4 



y=2sin x 




64. a = — |, b = |; f(x) — g(x) = 8 cos x — sec x 

(8 cos x — sec 2 x) dx = [8 sin x — tan x] _ 
= (8-f-V3)-(-8-#- 



/3 



6\/3 



y=8cosx 




(secx) 2 



* x 



65. a = -1, b = 1; f(x) - g(x) = (1 - x 2 ) - cos 



A: 



I J 



1 — X — COS I 



i)]dx 



2(i-i; 



4 _ 4 

3 7T 



y=1-x2 




66. A = Al + A2 

ai = — 1, bi = and a 2 = 0, b2 = 1; 

fi(x) - gi(x) = x - sin (f ) and f 2 (x) - g 2 (x) = sin (f ) - x 
=>• by symmetry about the origin, 

Ai + A 2 = 2Ai => A = 2 J [sin (f ) - x] dx 

, =2[(-f"0-i)-(-f-l-0)] 



2 I- 2 cos (^ 



V Z7T 



y=sin(*x/2) — 




i — x 



67. a = — f , b = |; f(x) - g(x) = sec 2 x - tan 2 x 

/ 7r/4 
(sec 2 x — tan 2 x) dx 

- /. 



[sec 2 x — (sec 2 x — 1)] dx 



?r/4 L 

/i-dx [xl;, 



7T/4 

/ 
-7r/4 



,7T/4 




68. c = - |, d = |; f(y) - g(y) = tan 2 y - (- tan 2 y) = 2 tan 2 y 
= 2(sec 2 y- 1) => A= I 2(sec 2 y-l)dy 

*^ -7r/4 



W 4 

4 - 7T 



2[tany-y]! / 7r 4 /4 =2[(l 
4fl — ? 



^/4 

-jt/4' 



•i + !)] 




x=(tany) 2 

i — x 



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340 Chapter 5 Integration 



69. c = 0, d = |; f(y) — g(y) = 3 sin y^/cos y — = 3 sin y^/cos y 

=> A = 3 J" sin y ycos^ dy = -3 [| (cos y) 3 / 2 ] ^ 
= -2(0 - 1) = 2 



70. a = -1, b = 1; f(x) - g(x) = sec 2 (^ 






3 



V3] dX 
3 



-tan I 

6^3 



t* 4/3 ]-i 




1- x 



71. A = Aj +A 2 

Limits of integration: x = y 3 and x = y 

=> y 3 - y = =>• y(y - l)(y + 1) = 
and c 2 = 0, d 2 



y = y° 

ci = — 1, di 



f2(y) - g2(y) = y 

Ai + A 2 = 2A 2 : 



i;fi(y)-gi(y) = y -y and 

— y 3 => by symmetry about the origin, 

* A = 2f'(y-y 3 )dy = 



n 



2(1 



l 




72. A = Ai + A 2 

Limits of integration: y = x 3 and y = x 5 







x 3 (x - l)(x + 1) = 



X' = x u 
» ai = -1, bi 



and a 2 = 0, b 2 = 1; f x (x) - g x (x) 



x 5 and 



f2(x) - g 2 (x) = J 

Ai + A 2 = 2A 2 



by symmetry about the origin, 



A 



'/.V 



dx 



2 a 




-i# 




73. A = Ai + A 2 

Limits of integration: y = x and y = \ =>- x=^,x^0 
=4> x 3 = 1 => x = 1 , fi(x) - gi(x) = x - = x 

=, Al = /Jx dx = [f ] * = I; f 2 (x) - g 2 (x) = 4, - 

=*- 2 ^ A 2 = jV 2 dx=[^] 2 = - 



1 



A = Ai + A 2 



2. 1/ = 1/x 




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Section 5.6 Substitution and Area Between Curves 341 



74. Limits of integration: sin x = cos x => 
and b = | ; f(x) — g(x) = cos x — sin x 

>?r/4 



f => a = 



pTT/4 

A = / (cos x — sin x) dx = [sin x + cos x] 

u 



tt/4 




V2 , V2 



-(0 + 1) = a/2-1 




75. (a) The coordinates of the points of intersection of the 

line and parabola are c = x 2 =>- x = ± yfc and y = c 
(b) f(y) — g(y) = y/y — (— -t/y) — 2,/y => the area of the 

lower section is, A L = I [f(y) — g(y)] dy 

= 2 J y/y dy = 2 [| y 3 / 2 ] C Q = f c 3 / 2 . The area of the 
entire shaded region can be found by setting c = 4: A = ft) 4 3//2 




, , , _ ^r = T • Since we want c to divide the region 



into subsections of equal area we have A = 2A L 



32 
3 



2(fc 3 / 2 ) 



12/3 



(c) f(x) - g(x) = c - x 2 => A L = J ^ [f(x) - g(x)] dx = / (c - x 2 ) dx = [ex - 



2, ,1, - .... vl ^ =2 |c 3 / 2 - c3: 



= I c 3 ' 2 . Again, the area of the whole shaded region can be found by setting c = 4 =>■ A 
condition A = 2A L , we get | c 3/2 = y => c = 4 2/3 as in part (b). 



y . From the 



76. (a) Limits of integration: y = 3 — x 2 and y = — 1 



-1 



a 



-2 and b 



f(x)-g(x)=(3-x 2 )-(-l) = 4-x 
=$► A = f (4 - x 2 ) dx = Ux - 



16 



-2 
16 — 32 
3 — 3 



(b) Limits of integration: let x = in y = 3 — x 2 

=> y = 3; f(y) - g(y) = v / 3 T ~y - (- x/3^7) 
= 2(3 - y) 1 / 2 

=> A = 2 J (3 - y) 1/2 dy = -2 J (3 - y) 1/2 (-l) dy = (-2) 



!/ = -! 



2(3 -y) 3 
3 




i)(8) 



32 
3 



[0 - (3 + l) 3 / 2 ] 



77. Limits of integration: y = 1 + ^Jx and y = -j- 

=> l + v /^=2^ x _^ ^ v /^ + x = 2 => x = (2-x) 2 

=> x = 4 - 4x + x 2 =>• x 2 -5x + 4 = 
=4> (x - 4)(x - 1) = => x=l,4 (but x = 4 does not 
satisfy the equation); y = A- and y = | =>• -j- = | 



x^/x 



64 



Therefore, AREA = Aj + A 2 : fi(x) - gl (x) = (l 



,1/2 



)-! 



* - 1: 






.1/2 



dx 



2-8/3 



y = l + v/5 




;i + i-i)-o 



g;f 2 (x)-g 2 (x) = 2x- 1 /2_| 



A 2 = /"(2X- 1 / 2 - |) dx = [4xV 2 - £] 



(4-2 



-(4-|)=4-f 



f ; Therefore, AREA = A : + A 2 = g 



n _ 37+51 
8 24 



88 
24 



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342 Chapter 5 Integration 



78. Limits of integration: (y — l) 2 = 3 — y => y 2 — 2y + 1 

= 3 - y => y 2 - y - 2 = => (y- 2)(y + 1) = 

=4> y = 2 since y > 0; also, 2,/y = 3 — y 

=> 4y = 9 - 6y + y 2 => y 2 - lOy + 9 = 

=$■ (y — 9)(y — 1) = =>• y=l since y = 9 does not 
satisfy the equation; 
AREA = Ai + A 2 

fi(y)-gi(y) = 2 v ^-o = 2y 1 / 2 

=> A 1 = 2 / o 'y 1/2 dy = 2 [^] * = f ; f 2 (y) - g 2 (y) = (3 - y) - (y - l) 2 

=* A 2 = / 2 [3 - y - (y - l) 2 ] dy = [3y - \ y 2 - i (y - l) 3 ] \ = (6 - 2 - |) - (3 - \ + 0) = 1 - § + 1 
Therefore, Aj + A 2 




_4,7_15_5 
v 2 -3 + 6-^-2 



79. Area between parabola and y = a 2 : A = 2 I (a 2 — x 2 ) 

Area of triangle AOC: \ (2a) (a 2 ) = a 3 ; limit of ratio = lim 



dx = 2 [a 2 x - 1 x 3 ] a Q = 2 (a 3 - f ) 



0=f; 



| which is independent of a. 



nb nb r*b r*b nb 

80. A = J 2f(x) dx - J f(x) dx = 2 J f(x) dx - J f(x) dx = J f(x) dx = 4 

81. Neither one; they are both zero. Neither integral takes into account the changes in the formulas for the 
region's upper and lower bounding curves at x = 0. The area of the shaded region is actually 

pO pl pl) pi 

A = J [-x - (x)] dx + J [x - (-x)] dx = J -2x dx + J 2x dx = 2. 



82. It is sometimes true. It is true if f(x) > g(x) for all x between a and b. Otherwise it is false. If the graph of f 
lies below the graph of g for a portion of the interval of integration, the integral over that portion will be 
negative and the integral over [a, b] will be less than the area between the curves (see Exercise 53). 

83. Let u = 2x => du = 2 dx => | du = dx; x=l =>• u = 2, x = 3 =>• u = 6 
J §2^5 dx = J^ stan (i du ) = J a» du = [F(u)] I = F(6) - F(2) 

84. Let u = 1 - x =>• du = -dx =4> -du = dx;x = => u=l,x=l =4> u = 

pl p0 pi) pi pi 

J f(l-x)dx=J f(u)(-du) = -J f(u)du = J f(u)du=J f(x) dx 

85. (a) Let u = -x =^ du = - dx; x = -1 =4> u = 1, x = =>■ u = 

/0 n0 pa no pl 

f(x)dx=J f(-u)(-du) = J -f(u)(-du) = J f(u)du=-J f(u) du 

= -3 
(b) Letu = -x => du = - dx; x = -1 =>- u = 1, x = => u = 

p0 pl) pO pl 

feven => f(-x) = f(x). Then J f(x) dx = J f(-u)(- du) = - J f(u) du = J f(u) du = 3 

86. (a) Consider J f(x) dx when f is odd. Let u = — x => du = — dx =>• — du = dx and x = — a =>• u = a and x = 

/0 pl) pO pa pa 

f(x) dx = J -f(-u) du = J f(u) du = - J f(u) du = - J f(x) dx. 

Thus J f(x) dx = J f(x) dx + J f(x) dx = - J f(x) dx + J f(x) dx = 0. 

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->-/2 
t 

-Ttd 



r 

(b) J sin x dx = [—cos x] ' , cos 



n 7T/2 

-ir/2 



Section 5.6 Substitution and Area Between Curves 343 
s(f)+cos(-f) =0 + = 0. 



87. Let u = a — x =>- du=— dx; x = =4* u = a, x = a =4> u = 

T _ P' f(x)dx _ P f(a-u) , .x _ P' f(a-u)du _ P" f(a-x) dx 
Jo «x)+f(a-x) J a f(a-u)+f(u) ^ UU ' J f(u)+f(a-u) J f(x)+f(a-x 



-x) 



Therefore, 21 = a =4> I 



2 ' 



Let u = S =>. du = - S dt => - — du = 7 dt => - - du = 7 dt; t = x => u = y, t = xy 

t v- xy t u t ' j •> j 

("■y pi pi py py 

I I dt = / - i du = - 1 du = / i du = I - dt 

Jx t Jy U J U J| U J| t 



u = 1. Therefore, 



Let u = x + c =4> du = dx; x = a — c => u = a, x = b — c =>• u = b 

f»b— c r»b nb 

I f(x + c) dx = I f(u) du = I f(x) dx 

l/ a— c *J a «-/ a 



90. (a) 



f(x+1)-(x + 1) 




(b) 



y f(x)-slnx 



(c) 



t-x 




f(x + |).sin(x + |) 




91-94. Example CAS commands: 
Maple : 

f := x -> x A 3/3-x A 2/2-2*x+l/3; 

g := x -> x-1; 

plot( [f(x),g(x)], x=-5..5, legend=["y = f(x)","y = g(x)"], title="#91(a) (Section 5.6)" ); 

ql:=[-5,-2, 1,4]; # (b) 

q2 := [seq( fsolve( f(x)=g(x), x=ql[i]..ql[i+l] ), i=l..nops(ql)-l )]; 

for i from 1 to nops(q2)-l do # (c) 

area[i] := int( abs(f(x)-g(x)),x=q2[i]..q2[i+l] ); 

end do; 

add( area[i], i=l..nops(q2)-l ); # (d) 
Mathematica : (assigned functions may vary) 

Clear[x, f, g] 

f[x_] = x 2 Cos[x] 

g[x_] = x 3 - x 

Plot[{f[x],g[x]}, {x, -2,2}] 
After examining the plots, the initial guesses for FindRoot can be determined. 

pts = x/.Map[FindRoot[f[x]==g[x],{x, #)]&, {-1, 0, 1}] 

il=NIntegrate[f[x] - g[x], {x, pts[[l]], pts[[2]]}] 

i2=NIntegrate[f[x] - g[x], {x, pts[[2]], pts[[3]]}] 

il +i2 



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344 Chapter 5 Integration 
CHAPTER 5 PRACTICE EXERCISES 

1. (a) Each time subinterval is of length At = 0.4 sec. The distance traveled over each subinterval, using the 

midpoint rule, is Ah = \ (v ; + v i+1 ) At, where v ; is the velocity at the left endpoint and v i+1 the velocity at 
the right endpoint of the subinterval. We then add Ah to the height attained so far at the left endpoint v, to 
arrive at the height associated with velocity v i+1 at the right endpoint. Using this methodology we build 
the following table based on the figure in the text: 



t (sec) 





0.4 


0.8 


1.2 


1.6 


2.0 


2.4 


2.8 


3.2 


3.6 


4.0 


4.4 


4.8 


5.2 


5.6 


6.0 


v (fps) 





10 


25 


55 


100 


190 


180 


165 


150 


140 


130 


115 


105 


90 


76 


65 


h(ft) 





2 


9 


25 


56 


114 


188 


257 


320 


378 


432 


481 


525 


564 


592 


620.2 



t (sec) 


6.4 


6.8 


7.2 


7.6 


8.0 


v (fps) 


50 


37 


25 


12 





h(ft) 


643.2 


660.6 


672 


679.4 


681.8 



NOTE: Your table values may vary slightly from ours depending on the v-values you read from the graph. 
Remember that some shifting of the graph occurs in the printing process. 
The total height attained is about 680 ft. 
(b) The graph is based on the table in part (a). 



h(i 


eet) 










, > 










700 












600 












500 












400 












300 












200 












100 















2 


4 


6 


8 





* t (sec) 



2. (a) Each time subinterval is of length At = 1 sec. The distance traveled over each subinterval, using the 
midpoint rule, is As = I (vj + v i+1 ) At, where v ; is the velocity at the left, and v i+1 the velocity at the 
right, endpoint of the subinterval. We then add As to the distance attained so far at the left endpoint v ; 
to arrive at the distance associated with velocity v i+1 at the right endpoint. Using this methodology we 
build the table given below based on the figure in the text, obtaining approximately 26 m for the total 
distance traveled: 



t (sec) 





1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


v (m/sec) 





0.5 


1.2 


2 


3.4 


4.5 


4.8 


4.5 


3.5 


2 





s(m) 





0.25 


1.1 


2.7 


5.4 


9.35 


14 


18.65 


22.65 


25.4 


26.4 



(b) The graph shows the distance traveled by the 
moving body as a function of time for 
< t < 10. 




10 , 10 
(a) E | = \ E a t = \ (-2) = - \ 
k = 1 k = 1 




10 10 

(b) E (b k -3a k )= E K- 

k= 1 k= 1 


10 

-3 E a 

k= 1 


10 10 10 

(c) E (a k + b k -l)= E a t + E b t - 

k= 1 k= 1 k= 1 


10 

- E i = 

k= 1 


= -2 + 25-(l)(10) = 13 





25 - 3(-2) = 31 



Copyright (c) 2006 Pearson Education 




Chapter 5 Practice Exercises 345 



(d) £ (§-W)= £ |- £ b k = |(10)-25 = 

k= 1 k= 1 k = 1 



20 20 

4. (a) £ 3a k = 3 £ a k = 3(0) = 
k= 1 k= 1 

20 20 20 

(c) £ (3-^) = £ |-7 S b, = i (20) - f (7) 
k=l k=l k=l 

20 20 20 

(d) £ (a* -2)= £ a t - £ 2 = - 2(20) = -40 
k= 1 k= 1 k= 1 



20 20 20 

(b) £ (a k + b t )= £ a k + £ b k = + 7 = 7 
k= 1 k= 1 k= 1 



5. Let u = 2x - 1 =4> du = 2 dx =>- \ du = dx; x = 1 => u = 1, x = 5 => u = 9 

£(2x - 1)+ 2 dx = £u+ 2 (i du) = [U 1 / 2 ] J = 3 - 1 = 2 

6. Let u = x 2 - 1 => du = 2x dx =4> i du = x dx; x = 1 => u = 0, x = 3 => u = 8 

£x (x 2 - 1) 1/3 dx = JV/ 3 (1 du) = [| u 4 / 3 ] J = I (16 - 0) = 6 

7. Let u = § => 2du = dx;x = -7r => U = - | , x = => u = 

J cos (!) dx = J (cos u)(2 du) = [2 sin u]^ /2 = 2 sin - 2 sin (- §) = 2(0 - (- 1)) = 2 



Let u = sin x =>■ du = cos x dx; x = => u = 0, x = f =4- u=l 



Jo 



(sin x)(cos x) dx 



Jo 



du 



9. (a 

(c 
(c 

10. (a 

(c 
(c 



J f(x) dx = | J 3 f(x) dx = 1 (12) = 4 (b) J f(x) dx = J f(x) dx - J f(x) dx = 6 - 4 = 2 



/; 2 g ( x) dx = - £ 



g(x) dx = - / g(x) dx 



(d) J (-■ 7r g(x)) dx = -7r J g(x) dx = -7r(2) = -2tt 



£(«+lW) dx= l£f(x)dx+l£g(x)dx=I(6) + i 



(2) 



/ o g(x) dx = i J o 7 g(x) dx = i (7) = 1 (b) £g(x) dx = J o g(x) dx - f o g(x) dx = 1 - 2 = - 1 



£f ( x)dx = -£f( 



x) dx = — IT 



(d) f \/2 f(x) dx = \fl \ f(x) dx = V^OO = tt^ 

*/ t/ 



f [g(x) - 3 f(x)] dx = f g(x) dx - 3 f f(x) dx = 1 - 3tt 

*J */ t/ o 



11. x 2 -4x + 3 = => (x-3)(x- 1) = => x = 3 orx= 1; 
Area = J (x 2 - 4x + 3) dx - J (x 2 - 4x + 3) dx 

f - 2x 2 + 3x1 - [y - 2x 2 + 3x] 

(^ -2(1) 2 + 3(1)) -O] 

- [(f-2(3) 2 + 3(3))-(f-2(l) 2 + 3(l))] 

(l + i)- [o- (1 + 01 = | 



f (x) = x - 4x + 3 




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12. l-^-=0^4-x 2 -0^x=±2 



346 Chapter 5 Integration 

1-^=0 => 4 - x 2 - => 

Area =£( i -f)* t -/ 4 3 ( i -f)* t 

(3-fi) 



2- £ 

^ 12 






n _ 4\ = B 

U 3^ 4 



f(x) = 1 - (x Z /4) 




13. 5 - 5x 2 / 3 = => 1 - x 2 / 3 = => x = ± 1; 

Area = J (5 - 5x 2 / 3 ) dx - J (5 - 5x 2 / 3 ) dx 

= [5x - 3x 5 / 3 ] ^ - [5x - 3x 5 / 3 ] I 
= [(5(1) - 3(1) 5 / 3 ) - (5(-l) - 3(-l) 5 / 3 )] 
- [(5(8)-3(8) 5 / 3 )-(5(l)-3(l) 5 / 3 )] 
= [2 - (-2)] - [(40 - 96) - 2] = 62 



J f(x) = 5 - 5X 273 




\2 4 6 8 


-5 


\. 


-10 


^N. 


-15 


^-s 



14. 1 - yx = => x= 1; 

Area = £ (l - ^x) dx - f*(l ^/i) dx 

= [ X _| X 3/ 2] l_ [x _| x 3/ 2] 4 

= [(1 _ | (1)3/2) _ ] „ [(4 _ 2 (4) 3/2) _ (1 _ 2 (1) 3/ 2) ] 

= i-[(4-¥)-J]=2 

15. f(x) = x, g(x) = 4j, a = 1, b = 2 => A = f [f(x) - g(x)] dx 

=j; 2 (x-i)dx=[f + ^=(i + i)-(i+i) = i 



16. f(x) = x, g(x) = -i=, a =l,b = 2 => A= f [f(x) - g(x)] dx 



J!V*) dx =[^2v^ 



f-2^2 



:i-2) 



7-4V2 



f(x) = 1 -Vx 





5 


f 








2 

1- 


y=x^ 




tllli 

y=VV 


X 






i 




2 



17. f(x) = (1 - ,/x) 2 , g(x) = 0, a = 0, b = 1 ^ A = J[ [f(x) - g(x)] dx = J q (l - y^) 2 dx = £ (l - 2,/x" + x) 



dx 



J 



'l-2x 1 / 2 + x)dx=: |x-fx 3 / 2 + ^ 



4 + I = I 
3 > 2 6 



(6-8 + 3) 



r*b r»\ nl 

18. f(x) = (1 - x 3 ) 2 , g(x) = 0, a = 0, b = 1 => A = J [f(x) - g(x)] dx = J (1 - x 3 ) 2 dx = J (1 - 2x 3 + x e 



dx 



X I X 



1 ' 7 | 1 2 ' 7 14 



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Chapter 5 Practice Exercises 347 



19. f(y) = 2y 2 , g(y) = 0, c = 0, d = 3 

=>• A = J [f(y) - g(y)] dy = J q (2y 2 - 0) dy 



r 



y dy = I [y ] 



18 




*- x 



20. f(y) = 4 - y 2 , g(y) = 0, c = -2, d = 2 

=* A = f [f(y) - g(y)] dy = £(4 - y 2 ) dy 



4y 



2(8 



32 



x=4-y 2 




21. Let us find the intersection points: ^- = %±- 

=> y 2 - y - 2 = => (y- 2)(y + 1) = => y 



or y 



■1, d = 2; f(y) = y-f , g(y) 



A = f [f(y) - g(y)] dy = £ (*±* - £ ) dy 

f £ l (y + 2-y 2 )dy=i[£+2y-f '"' 



H(i 



4 L I 2 + 4 3 ; 



^2 z ^ 3 



I)] 




22. Let us find the intersection points: ^— = y ^ 

=> y 2 - y - 20 = => (y - 5)(y + 4) = => y = -4 



y + 16 



y 2 -4 



ory = 5 =* c = -4, d = 5; f(y) = ^ , g(y) - A 
=* A = £ [f(y) - g(y)] dy = J^^ 16 - ^) dy 



5£(y + 2 °-y 2 ) d y = 3p + 20y-f 

I[(f + 100-f)-(f-80+f)] 



4 
1 f9 

4 



(1 + 180-63) 



1 /9 



117) = i(9 + 234) 



243 




x= (y+16)/4 



23. f(x) = x, g(x) = sin x, a = 0, b = | 

n b p 7r/4 

I [f(x) — g(x)] dx = I (x — sin x) dx 

t./ >i ' Jo 

I 7I "/ 4 / 2 



A 



+ «,]r-(s + ^) 



i 



y=x 



y=sinx 



-* — x 



jt/4 



- b = - 

2' u 2 



24. f(x)= l,g(x)= |sinx|,a 

=>• A = f [f(x) - g(x)] dx = f^ (1 - |sin x|) dx 

i/ a t/ — 7r/2 

(1 + sin x) dx + I (1 — sin x) dx 
= 2 / (1 — sin x) dx = 2[x + cos x] ' 



2(1-0 



7T-2 



Copyright (c) 2006 Pearson Education 




1 — X 



348 Chapter 5 Integration 



25. a = 0, b = 7r, f(x) — g(x) = 2 sin x — sin 2x 
=>- A = I (2 sin x — sin 2x) dx = [—2 cos : 
= [_2.(-l)+i]-(-2-l + I)=4 



cos 2x 1 ^ 




26. a 



|, b = |, f(x) — g(x) = 8 cos x — sec 2 x 
c - 
(-8 • ^ + a/3) = 6x/3 



pT/3 ^g 

A = / (8 cos x — sec 2 x) dx = [8 sin x — tan x.] 71 ' '' 

*/ — 7T/3 



tt/3 



2 



y=8cosx 




=(secx) 2 
i x 



27. f(y) = ^y, g(y) = 2-y,c = l,d = 2 

=► A = / [f(y) - g(y)] dy = J] [^/y - (2 - y)] dy 

= / 2 (^-2 + y)dy= [| y 3 / 2 - 2y + 



| V 2 - 4 + 2 



;i-2+i; 



7 _ 8\/2-7 
6 6 




28. f(y) = 6 - y, g(y) = y 2 , c = 1, d = 2 
=> A = j [f(y) - g(y)] dy 

= [ey-*-*]'=(i2-2-| 

= 4 



/ 2 (6 - y - y 2 ) dy 



7 , 1 _ 24-14+3 
3^2 6 



13 

6 



1- 



y=VL 



r=6-x 



-i — X 



4 5 



29. f(x) = x 3 - 3x 2 = x 2 (x - 3) =>• f (x) = 3x 2 - 6x = 3x(x - 2) =$■ f = +++ | | ++- 

2 

=>• f(0) = is a maximum and f(2) = —4 is a minimum. A = — / (x 3 — 3x 2 ) dx = — $ — j 



27 
4 



30. A 



(f-27) 

JJ(a 1/2 - XV2) 2 dx = £{a - 2 v ^x 1 /2 + x ) dx = [ax - \ y/i: 
f (6 - 8 + 3) = f 



3/2 I jr 



2, 1 1 



3^2; 



— a — j -y/a • a^a 



3 1 . The area above the x-axis is Ai = I (y 2//3 — y) dy 



jq ; the area below the x-axis is 



£<> 



2/3 _ 



y) dy 



3y°/' 3 
5 



11 

10 



the total area is Ai + A 2 




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Chapter 5 Practice Exercises 349 



X7r/4 n 57r/4 

(cos x — sin x) dx + I (sin x — cos x) dx 



J 5tt/4 



(cos x — sin x) dx = [sin x + cos x] 



[- cos x - sin x] ,' 4 + [sin x + cos x] 57r y 4 



7T/4 



y = sin i 




= [(f + #) -(o+i)] + [{4 + 4)-{-4-4)} 

+ [(-1+0)- (-4-4)} = *~4 -2 = 4^2-2 



33. y = x 2 + f*\ dt ={► | = 2x + 1 => g = 2 - + ; y(l) = 1 + £ \ dt = 1 and y'(l) = 2+1=3 

34. y = J (1 + 2^/sect) dt =>• & = 1 + 2^/sec x =>• = 2 Q) (sec x) _1 / 2 (sec x tan x) = ^/sec x (tan x); 

(l + 2^/sec t) dt = and x = =4> 

35. y= J^dt-3 => g = ^;x = 5 => y = //** 

36. y = J \/2 - sin 2 1 dt + 2 so that ^ = y/l - sin 2 x; x = - 1 => y = J \f 



% = l + 2v^0 = 3 



dt - 3 = -3 



2 - sin 2 t dt + 2 = 2 



37. Let u = cos x =>• du = —sin x dx =>• — du = sin x dx 

J 2(cos x)-V2 S i n x dx = J 2iT 1 / 2 (- du) = -2 J iT 1 / 2 du = -2 (j?£) + C = -4U 1 / 2 + C 
= -4(cos x) 1 / 2 + C 



38. Let u = tan x =>• du = sec - x dx 



J (tan x)- 3 / 2 sec 2 x dx = J u" 3 / 2 du = f^ + C = -2U- 1 / 2 + C = j^]- 2 + C 



39. Let u = 26> + 1 => du = 2 d0 =5> § du = d0 

J [20 + 1 + 2 cos (20 + 1)] d0 = J (u + 2 cos u) (± du) = f + sin u + Ci - , 
= 2 + + sin (20 + 1) + C, where C = Ci + J is still an arbitrary constant 



2=t^-Lsin(20+ 1) + Ci 



40. Let u = 20 - 7T => du = 2 d0 => ± du = d0 



/(y3b + 2sec2 ^- 7r >) d0 = /(7n + 2sec2u )(5 du ) = 5/K 1/2 + 2 

= \ (^-\ + \ (2 tan u) + C = u 1 / 2 + tan u + C = (20 - tt) 1 / 2 + tan (20 - tt) + C 



sec" u) du 



J< 



41. I (t-|) (t-Hf)dt: 



J(t 2 -|)dt=/(t 2 -4r 2 )dt=f-4(^)+C=f + t+C 



42. 



J^^dt=/^dt=J, 



P + ?)* : 



/o 



r 2 + 2r 3 ) dt : 



(-i) 



(S) 



+C=-i- 



i - i + C 

t f2 + *" 



43. Let u = 2t 3 / 2 => du = 3^ldt =>• |du = i/t dt 

J ^/tsin (2t 3 / 2 )dt = 1 J sin u du = -jcos u + C = -±cos(2t 3 / 2 ) + C 



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350 Chapter 5 Integration 



44. Let u = 1 + sec 6 => du = sec 9 tan# 69 =>■ J sec 6 tmO \/l + sec 9 69 = J u^du = §u 3 / 2 + C 



§(l + sec6>) 3/2 +C 



45. 



46. 



47. 



J (3x 2 - 4x + 7) dx = [x 3 - 2x 2 + 7x] ^ = [l 3 - 2(1) 2 + 7(1)] - [(-l) 3 - 2(-l) 2 + 7(-l)] = 6 - (-10) = 16 
J q (8s 3 - 12s 2 + 5) ds = [2s 4 - 4s 3 + 5s] J = [2(1) 4 - 4(1) 3 + 5(1)] -0 = 3 

/V 4 / 3 dx = [-3X- 1 / 3 ] J 7 = -3(27)^/3 - (-3d)- 1 / 3 ) = -3 (§) + 3(1) = 2 

r*=r^=r r>/8 *=[- 2ri/s ]j 



-2 _ (-2) 



50. Let x = 1 + Ju => dx = \ ir 1 / 2 du => 2dx=-^=;u=l => x = 2, u = 4 => x = 3 

J' ii± ^ ! du = £ xV2( 2 dx) = [2 (|) x 3 / 2 ] J = 4 (3 3 / 2 ) - I (2 3 / 2 ) = 4^3 - | v^ = f (3 v/3 - 2^2 



51. Letu = 2x+ 1 =>• du = 2 dx =>• 18 du = 36 dx; x = => u= l,x = 1 => u = 3 



r^ = r ^ du = w ! = ^ i = (^ - (t*) 



52. Let u = 7 - 5r =>■ du = -5 dr =4> - i du = dr; r = =^> u = 7, r = 1 =>• u = 2 

£ 1/ ^ W = £ (7-5rr 2 / 3 dr=/ 7 2 u- 2 / 3 (-Idu) = -I[3u 1 / 3 ] 2 7 =|( 3 v ^- 3 v ^ 



53. Let u = 1 - x 2 / 3 => du = - | x" 1 / 3 dx => - | du = x" 1 / 3 dx; x = | =>• u = 1 - (|) 2/3 = | 



1 => u = 1 - l 2 / 3 = 



£ x " /3 ( J - x2/3 ) 3/2 dx = C u3/2 ( " i du) = [ ( - ^ (f )1 L = [" i u5/2 ] 



1/8 

27^ 
160 




3/4 



3 ( )5/2 



!) G) 5/2 



54. Let u = 1 + 9x 4 => du = 36x 3 dx => ^ du = x 3 dx; x = => u = 1, x = ± => u = 1 + 9 Q) 4 

J o 1/2 X 3 (1+9xr 3/ 2dx= f /16 U -3/ 2 (l du ) = [1 (^)]; 5/16 = r X _-!/»! -A- 

= _ J, (25\-l/2 _(_±. (l)-l/2) = J_ 
18 U6^ V 18 KL > I 90 



n 4 _ 25 

16 



J_ -1/2] 25 
18 u J 1 



55. Let u = 5r => du = 5 dr =4> j du = dr; r = =>■ u = 0, r = 7r => u = 57r 

£sin 2 5rdr= J"(sin 2 u) (§ du) = § [§ - ^] ^ = (| - ^) - (0 - ^) = f 



56. Let u = 4t - | =4> du = 4 dt =>• j du = dt; t = =^ u 



3tt 



£t=-=>U— - 
4 ' l 4 ^ u 4 



p?r/4 /»37r/4 

/ cos 2 (4t-|)dt= I (cos 2 u)(idu) = i[ 

t/ <-/ — 7T/4 



i [u , sin2u ] 37r /4 _ i /^ sin(f) \ _ 1 / tt , sin (- |) \ 
4 L2 " r 4 J -tt/4 ~~ 4 ^ 8 T 4 J 4 \ 8' 4 ^ 



7T J_ I J_ ZI 

8 16 ' 16 ~~ 8 



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Chapter 5 Practice Exercises 35 1 



nw/3 

57. J sec 2 6d9 = [tan 8] n < 3 = tan f - tan = \fl 



58. 



J»37r/4 
C 
rr/4 



esc 2 x dx = [-cot xfj' = (- cot &) - (- cot |) = 2 



59. Let u = | =>- du = ^ dx =>• 6 du = dx; x = 7r => u = g, x = 3n =>- u = | 

DO o 2 



J cot 2 fdx=J 6cot 2 udu = 6j (csc 2 u - 1) du = [6(-cot u - u)]^g = 6 (- cot f - f) - 6 (-cot | - f) 



6^3 -2tt 



60. Let u = | =^ du = ± d(9 =>■ 3 du = &6; 9 = =>• u = 0, 6» = tt => u = f 



J tan 2 fd6>=J (sec 2 f - l) dff = J 3 (sec 2 u - 1) du = [3 tan u - 3u]£ 
= [3 tan f - 3 (f )] - (3 tan - 0) = 3a/3 - tt 



tt/3 



61. 



I sec x tan x dx = [sec x]^ , 3 = sec — sec (— |) = 1 — 2 = — 1 



-?r/3 
>3tt/4 



J»3 
csc z cot z dz = [—esc z]J){ = f— esc 3 ' ^ 
7T/4 



J 7t/4 



csc I) = -V2- 



'2 = 



63. Let u = sin x => du = cos x dx; x = => u = 0, x = f =$■ u=l 

■>jr/ 2 



J 5(sin x) 3 / 2 cos x dx = J 5u 3 / 2 du = [5 (|) u 5 / 2 ] Q = [2u 5 / 2 ] Q = 2(1) 5 / 2 - 2(0) 5 / 2 = 2 
64. Let u = 1 - x 2 => du = -2x dx => - du = 2x dx; x = - 1 => u = 0, x = 1 => u = 



i: 



2x sin (1 — x ) dx 



I' 



— sin u du = 



65. Let u = sin 3x => du = 3 cos 3x dx =>• | du = cos 3x dx; x 



f_ 15sin 4 3xcos3xdx= J 15u 4 (± du) = J 5u 4 du = [u 5 ] 1 : 



u = sin i 



&) = l,x=f => u = sin(f) 



(-1) 5 -(D 5 



66. Letu = cos (|) =4> du = - \ sin (|) dx => -2 du = sin (§) dx; x = =4> a = cos (§) = 1, x = ^f =3- u = cos ( -§-J 



l 

2 

"•27T/3 



// cos- 4 (|)sin(f)dx=j; "u- 4 (-2du)=[-2(^)]| 2 =| ( I)- 3 -|(i r 3 = 2 (8 _ 1)= i4 



67. Let u = 1 + 3 sin 2 x =>• du = 6 sin x cos x dx =>■ | du = 3 sin x cos x dx; x = =>• u=l,x=| 
=> u = 1 + 3 sin 2 |=4 

X V2 7ms dx = f * (i du ) = r s u " /2 du = [Kf)l ! = t ul/2 ] I = 4l/2 - il/2 = l 



68. Let u = 1 + 7 tan x =>■ du = 7 sec 2 x dx =>■ i du = sec 2 x dx; x = =>■ u = 1 + 7 tan = 1, 
x = | => u = 1 + 7 tan f = 8 

/r W f^*=r*(H«)=ri"- 2/s < i »=[Kf)]!=[? ui/! ']i=' <8) ' /3 -' (i > i/s =' 

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352 Chapter 5 Integration 



69. Let u = sec 9 => du = sec 9 tan 9 d$: 9 = =$ u = sec = 1, 



u = sec f=2 



/.ir/3 
Jo 



tan 



i/ 2 sec i 



d0 



Jo 5 



2 LC-i) 



sec 9 v 2 sec 

2 



d9 = r^^k dS = l 



1 du- -4- 



-s« 



3/2 du 



/2u 



J 1 n/2(2) I x/2Tl)J 



'2-1 



70. Let u = sin ^t => du = (cos 0) (± r 1 / 2 ) dt = ^ dt => 2 du = ^ dt; t = ^ 







6 2 



t = ^ => u = sin § = 1 



J ir 2 /36 



-^^ dt =j r i 2 7u( 2du )= 2 J r l u " /2du =[ 4 v / ^]: /2 = 4 ^ i - 4 yi= 2 ( 2 -^ 



yt sin x/t 



71. (a) av(f) = i^T) /_', (mx + b)dx = \ [s£ + bx] ^ = 1 [(sQ£+b(l)) - (=*=£+ b(-l))] = i (2b) = b 

(b) av(f) = ^ £ (mx + b) dx = £ [=£ + bx] ^ = £ [("f 2 + b(k)) - (=*=* + b(-k))] 
= £ (2bk) = b 

72. (a) y av = ^/^dx = \f ^^' 2 dx = f [| x 3 / 2 ] „ = f [f (3) 3 / 2 - § (O) 3 / 2 ] = f (2^) = 2 

(b) y« = ^0 T v^ dx = s/>* 1/2 dx = 4 [1 x3/2 ] o = ^(! ^ 3/2 - I (°) 3/2 ) = £ (I a v^) = 1 a 



J ^/axf '(x) dx = g^ [f(x)]J = ^ [f(b) - f(a)] = ^ff^ so the average value of f ' over [a, b] is the 
slope of the secant line joining the points (a, f(a)) and (b, f(b)), which is the average rate of change of f over [a, b]. 



73. f' = JT- 
av b — a 



74. Yes, because the average value of f on [a, b] is g-^ I f(x) dx. If the length of the interval is 2, then b — a = 2 
and the average value of the function is | I f(x) dx. 



75. We want to evaluate 

•*365 

305 



p365 />365 / 

s^oJo f W dx =355j 37sin 



l^-io 1 ) 



25 dx = J£ 



305 



r 

Jo 



sin 



2- 
505 



(x- 101) 



dx 



25 
305 



I' 

Jo 



dx 



Notice that the period of y = sin |^ (x — 101 



2^ 

575 



is -r = 365 and that we are integrating this function over an iterval of 



length 365. Thus the value of 



305 



Jo 



2- 
:in.-. 



(x-101) 



n365 

dx + H I dx is J£ ■ + H ■ 365 = 25. 

365 J 365 365 



76 - 675^20 /r^ 8 " 27 + 10_5 ( 26T - L87T2 )) dT = 6^5 



8.27T - 



26T 2 _ 1.87T 3 
2-10 5 3-10 5 



675 



1 
655 



.27(675) 



26(675f 1.87(675)-' 



2-10 5 



3-K^ 



07On\ -1- 26 ( 20 ) 2 _ 1 - 87 (2Q) 3 

■ Z 'W 1" 2-105 3 . 10 5 



20 

■ J « ^(3724.44 -165.40) 



= 5.43 = the average value of C v on [20, 675]. To find the temperature T at which C v = 5.43, solve 
5.43 = 8.27 + 10~ 5 (26T - 1.87T 2 ) for T. We obtain 1.87T 2 - 26T - 284000 = 

26±,/(26) 2 -4(1.87)(-284000) _ 2 6±y / 212 4996 



2(1.87) 

interval [20, 675], so T = 396.72°C. 



3.74 



-. So T = -382.82 or T = 396.72. Only T = 396.72 lies in the 



J> 



V^ 



77. g = \/ 'i + cos-'x 



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Chapter 5 Practice Exercises 353 



78. g = v/2 + cos 3 (7x 2 ) • |^(7x 2 ) = 14x^2 + cos 3 (7x 2 ) 



7Q dy _ _d_ 
ly - dx — dx 



Ji 3 + t 



dt 



6 
3+x 4 



80. | 



A fr 2 i 

dx \ J sec x t2 + 1 



dt 



/; 



p+i 



dt 



1 d / scc \ _ sec x tan x 

sec 2 x + 1 dx *■ * 1 + sec 2 x 



81. Yes. The function f, being differentiable on [a, b], is then continuous on [a, b]. The Fundamental Theorem of 
Calculus says that every continuous function on [a, b] is the derivative of a function on [a, b]. 

82. The second part of the Fundamental Theorem of Calculus states that if F(x) is an antiderivative of f(x) on 

nb 

[a, b], then I f(x) dx = F(b) — F(a). In particular, if F(x) is an antiderivaitve of y 1 + x 4 on [0, 1], then 
J q y4 +x 4 dx = F(l) - F(0). 



83. y 



84. y 



r 



1 + 1 2 dt = - 






1 + 1 2 dt 



dy _ _d_ 

dx dx 



/.x 



1 + t 2 dt 



d_ 

dx 



/iX 



1 + 1 2 dt 



-yTT^ 



dt 



r 



r^ dt 



dy _ 

dx dx 



r pcosx r pcosx 

E [~ J„ T^t2 dt = ~ dx" |y o ~ 



dt 



. 1 — cos 2 x 7 V dx 



E ( cos x >) 



t\-) (— sin x) = -J— = esc x 

m^ x / v y sin x 



85. We estimate the area A using midpoints of the vertical intervals, and we will estimate the width of the parking lot on each 
interval by averaging the widths at top and bottom. This gives the estimate 



Ass 15 • 



^ + 36 , 36 + 54 , 54 + 51 , 51 + 49.5 , 49.5 + 54 , 54 + 64.4 , 64.4 + 67.5 , 67.5 + 42 
^2" l "2~ t "2" t "2" r 2" ,_ 2 _r 2 '2 

>2 XKo ™ot io A^ao . f"<tO 1fl/ft2\ r- /"<QA1 ff2\ f<£1 in/fi-2\ 



') 



A«5961ft 2 .ThecostisArea-($2.10/ft 2 )« (5961 ft 2 ) ($2.10/ft 2 ) = $12,518.10 => the job cannot be done for $11,000. 



86. (a) Before the chute opens for A, a = —32 ft/sec 2 . Since the helicopter is hovering, Vo = ft/sec 

=> v = J -32 dt = -32t + v = -32t. Then s = 6400 ft => s = J -32t dt = -16t 2 + s = -16t 2 + 6400. 
At t = 4 sec, s = -16(4) 2 + 6400 = 6144 ft when A's chute opens; 

(b) For B, s = 7000 ft, v = 0, a = -32 ft/sec 2 =4> v = J -32 dt = -32t + v = -32t =4> s = J -32t dt 

= -16t 2 + s = -16t 2 + 7000. At t = 13 sec, s = -16(13) 2 + 7000 = 4296 ft when B's chute opens; 

(c) After the chutes open, v = - 16 ft/sec => s = J - 16 dt = - 16t + s . For A, s = 6144 ft and for B, 

s = 4296 ft. Therefore, for A, s = - 16t + 6144 and for B, s = - 16t + 4296. When they hit the ground, 







for A, 0= -16t + 6144 => t 



6144 
16 



384 seconds, and for B, = - 16t + 4296 => t 



4296 
16 



268.5 seconds to hit the ground after the chutes open. Since B's chute opens 58 seconds after A's opens 
> B hits the ground first. 



p30 

87. av(I) = i J (1200 - 40t) dt = i [1200t 



20t 



2]30 
Jo — 30 



^ [((1200(30) - 20(30) 2 ) - (1200(0) - 20(0) 2 )] 



±5 (18,000) = 600; Average Daily Holding Cost = (600)($0.03) = $18 



3(1 



/»14 

av(I) = n J (600 + 600t) dt = ^ [600t + 300t 2 ]J = ± [600(14) + 300(14) 2 - 0] = 4800; Average Daily 
Holding Cost = (4800)($0.04) = $192 



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354 Chapter 5 Integration 



89. av(I) = jof () (450 - I) dt = i f450t - f 1 = i [450(30) - ^ - o] = 300; Average Daily Holding Cost 
= (300)($0.02) = $6 



90. av(I) = i J o ^600 - 20-/l5t) dt = gg £ (600 - 20\/l5 l 1 / 2 ) dt = i [600t - 20^15 (§) t 3 / 2 
= ^ [600(60) - ^^ (6O) 3 / 2 - Ol = i (36,000 - (2|Q) 15 2 ) = 200; Average Daily Holding Cost 
= (200)($0.005) = $1.00 



CHAPTER 5 ADDITIONAL AND ADVANCED EXERCISES 

1. (a) Yes, because f f(x) dx = ± f 7f(x) dx = ± (7) = 1 

(b) No. For example, J o 8x dx = [4x 2 ] J = 4, but J] V8x dx = [2^ (^l)l = ^ (l 3 / 2 - : 



3/2) 



1 dx = — 3 



2. (a) True: J f(x) dx = - J f(x) 1 

(b) True: fjfQQ + g(x)] dx = /_,f(x) dx + J^gW dx = J_/(x) dx + J] f(x) dx + J_ ? g(x) dx 

=4+3+2=9 

(c) False: J f(x) dx = 4 + 3 = 7 > 2 = J g(x) dx => J [f(x) - g(x)] dx > => J*_ [g(x) - f(x)] dx < 0. 
On the other hand, f(x) < g(x) =>■ [g(x) — f(x)] > =£• I [g(x) — f(x)] dx > which is a contradiction. 



3. y = i f Q f(t) sin a(x - t) dt = lf g f(t) 
= ssls J"f(t) cos at dt - £21M J" f(t) 



sin ax cos at dt — 



dx 



cos ax sin at dt 
cos at dt 



sin ax / _d_ 
a I dx 



sin at dt 



d^v 
dx-' 



J f(t) cos at dt J + sin ax J^ f(t) sin at dt - seis* | A J f( t ) , 

cos ax J f(t) cos at dt + ^f^ (f(x) cos ax) + sin ax J f(t) sin at dt - ^^ (f(x) sin ax) 
> ^ = cos ax I f(t) cos at dt + sin ax I f(t) sin at dt. Next, 
= —a sin ax I f(t) cos at dt + (cos ax) I ~ I f(t) cos at dt ) + a cos ax I f(t) sin at dt 



(sin ax) 



J>' 



J> 



sin at dt = — a sin ax / f(t) cos at dt + (cos ax)f(x) cos ax 



+ a cos ax I f(t) sin at dt + (sin ax)f(x) sin ax = —a sin ax I f(t) cos at dt + a cos ax I f(t) sin at dt + f(x). 

•J Jo Jo 

Therefore, y" + a 2 y = a cos ax I f(t) sin at dt — a sin ax I f(t) cos at dt + f(x) 

^ J o f(t) cos at dt - £2|S J f(t) sin at dt J = f(x). Note also that y'(0) = y(0) = 0. 



4. x 



= SI 7^ * =* = (X) = ^IoV^ dt =r y [£ 7^ dt] (I) from the chain rule 
|(1 + 4yT 1/2 (8y) (g) = ^M = M^l = 4 y. Thus g = 4y, and the constant of 

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Chapter 5 Additional and Advanced Exercises 355 



proportionality is 4. 



5. (a) £ f(t) 

=► f(x 2 ) 

(b) I t 2 dt 

«7 



dt = X COS 7TX =>• ^ I f(t) dt = COS 7TX — 7TX sin 7TX =^> f (X 2 ) (2x) = COS 7TX — 7TX sill 7TX 

COS 7TX — 7TX Sin 7TX Tl-1.10 v O — ^ fY/1 \ COS 2tT — 27T SIB 2?T 1 



2x 



Thus, x = 2 

) 3 

- '?, i, ,. _ 3 



f(x) 

, I =5 (f(x)) 3 

■5 I 3 



«4) - 

\ (f(x)) 3 = X COS 7TX => (f(x)) 3 = 3x COS 7TX 



f(x) = VJ 



X COS 7TX 



=> f(4) = V 3(4) cos 4tt = \A2 
6. J f(x) dx = | + | sin a + § cos a. Let F(a) = J f(t) dt => f(a) = F'(a). Now F(a) = f + § sin a + § cos a 



| + | sin | 



=>■ f(a) = F (a) = a + ± sin a + | cos a - | sin a =4> f (| 
7. J] f(x) dx = x/b 2 + l - ^2 =► f(b) = & J^ f(x) dx = 1 (b 2 + l) _1/2 (2b) 



W- cos 5 - I sin s - - 



1 _ w _ 1 
2 2 ""* 2 2 ' 2 2 2 



x/b^TT 



f(x) 



vA^+T 



8. The derivative of the left side of the equation is: -jj- I I f(t) dt du = / f(t) dt; the derivative of the right 
side of the equation is: £■ I f(u)(x — u) du = ^ I f(u) x du — -jj- I u f(u) du 



H [*J> 



)du 



_d_ 
dx 



J u u f(u) du = J f(u) du + x \£ J f(u) du - xf(x) = j^ f(u) du + xf(x) - xf(x) 
= I f(u) du. Since each side has the same derivative, they differ by a constant, and since both sides equal 

J 

when x = 0, the constant must be 0. Therefore, I I f(t) dt du = I f(u)(x — u) du. 



9. !| = 3x 2 + 2 => y = J (3x 2 + 2) dx = x 3 + 2x + C. Then (1,-1) on the curve => l 3 + 2(1) +C = -1 =4> C = — : 

=> y = x 3 + 2x - 4 

10. The acceleration due to gravity downward is —32 ft/sec 2 =4- v = I —32 dt = — 32t + Vo, where Vo is the initial 

velocity =>• v = -32t + 32 => s = J (-32t + 32) dt = -16t 2 + 32t + C. If the release point, at t = 0, is s = 0, then 
C = => s = -16t 2 + 32t. Then s = 17 => 17 = -16t 2 + 32t => 16t 2 - 32t + 17 = 0. The discriminant of this 
quadratic equation is —64 which says there is no real time when s = 17 ft. You had better duck. 



n3 p0 ni 

11. I f(x)dx= / x 2 / 3 dx+ I -4dx 

J -8 J -8 

-4x] 3 



[l^ 3 ]^ 



(0 - | (-8) 5 / 3 ) + (-4(3) - 0) 



36 
5 



f -12 



dx 





y = x 2/1 


4 
2 








-8 


-4 







3 


X 






-4' 


L y = 


-4 
— • 





12. £f(x) dx = £^ dx + £ (x 2 - 4) 

= [-f(-^) 3/2 ]- 4 +[f-H[ 

(-l(4) 3 / 2 )] + [(f-4(3))-0] 



Lo-(- 

16 _ o _ Z 

3 J — 3 




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356 Chapter 5 Integration 



■ /. 


g(t) dt = 


•J 


t dt + 1 sin 7rt 


= 


[«:+ 


[-1 


1 2 
COS 7TtJ 




= 


(J-o) 


+ [- 


- - cos 2ir 


-(- 




1 2 

2 TT 




i 





dt 



14. J h(z) dz = J v 7 ! -z dz + J (7z - 6)" 1 / 3 dz 
= [-§(!- z)3/ 2 ] J + [^ (7z - 6)^3] J 
= [-|d- D 3/2 -(-Id- O) 3 / 2 )] 

"£ (7(2) - 6) 2 / 3 - i (7(1) - 6) 2 / 3 ] 



1 L 14 

2 , /"6 



6 _ i_) — 55 

7 14/ — 42 





15. J f(x) dx = J '_ dx + J (1 - x 2 ) dx + J 

= w:£ + [x - f ] ] i + [2x] 2 



(-l-(-2)) + 



(l-f)-(-l-^ 2 ) + [2(2) -2(1)] 



1 + 



l)+4-2=f 



16. J h(r) dr = J r dr + J (1 - r 2 ) dr + J dr 



,, + wi 



(0-^) + ((l-f)-0) + (2-l) 

-1+1+1=1 



y=l 



y=2 




v-1 




17. Ave. value = j^/Vd) dx = ji, £f(x) dx = 1 [/Jx dx + Jj'(x - 1) dx] = 1 [*] V 1 [$ - x] ' 

= »[(¥-o) + (*-2)-(f-i)] = j 

18. Ave. value = ^ J] f(x) dx = 3+0 £ f(x) dx = | |£ dx + Jj dx + £ dx = i [1-0 + + 3- 2] 



19. 



f «=£Td^fw=Hi)-(i)aa)) = ^-x(-^) = ^+Hi 



2a f W = £°Jr^ dt =* f '«=(HiK)(s( sinx ))-(i^^)(s( cosx )) 

cos x sin x 



cos x 1 sin x 

cos 2 x sin 2 x 



21. g(y) = / v . y sint 2 dt =► g'(y)= (sin(2 v ^) 2 ) (| ( 2> /y)) - (sin(^) 2 ) (|(^y)) 



sin 4y sin y 



22. f(x) = f + t(5 - t) dt => f (x) = (x + 3)(5 - (x + 3)) {i (x + 3)) - x(5 - x) (g) = (x + 3)(2 - x) - x(5 - x) 

= 6 - x - x 2 - 5x + x 2 = 6 - 6x. Thus f (x) = =4> 6-6x = =>• x=l. Also, f"(x) = -6 < =4> x = 1 gives a 

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Chapter 5 Additional and Advanced Exercises 357 



maximum. 



23. Let f(x) = x 5 on [0, 11. Partition [0, 1] into n subintervals with Ax = — = -. Then 1, -,...,- are the 

v J l j j l j j n n n ' n ' ' n 

oo / . \ 5 

right-hand endpoints of the subintervals. Since f is increasing on [0, 1], U = J2 \i) (n) * s me u PP er sum f° r 



f(x) = x 5 on [0, 1] => lim f] ( ±-Y (±) = lim ± [(i) 

- J'" 1 

Jo 



CD : 



lim 

n — > oo 



l° + 2^ + ... + ir 

p6 



5 dx - . . 
o L 6 J o 



24. Let f(x) = x 3 on [0, 1]. Partition [0, 1] into n subintervals with Ax 



1-0 



n n 



i Then ±, = , ... , i are the 



oo / ■ \ 3 

right-hand endpoints of the subintervals. Since f is increasing on [0, 1], U = J2 [ i ) („) * s me u PP er sum f° r 



f(x) = x 3 on [0,1] => lim 



oo / . \ 3 

E (i) (J) 



lim i (1} 

n — > cc n \\ n ) 



(2) 3 '| :_ Jim 1^ n - 



r'x 3 

Jo 



dx 



25. Let y = f(x) on [0, 1]. Partition [0, 1] into n subintervals with Ax = — - = -. Then -, -,...,- are the 

- 7vyL,J l:j nn n'n''n 

right-hand endpoints of the subintervals. Since f is continuous on [0, 1], £) f ( - ) Q) is a Riemann sum of 
y = f(x)on[0,l] => n lim^Ef(0(D= n lim o \ [f (±) + f (f) + •■ • + f (=)] = £ f(x) dx 

26. (a) Jjm^ ^[2 + 4 + 6+... + 2n] = n Em, 1 [2 + 4 + 5 + .. . + f] = J^x dx = [x 2 ]J = 1, where f(x) = 2x 

on [0, 1] (see Exercise 25) 

<W n^oo ^[l 15 + 2 15 + -+» 15 ] = n !H n oc ^[a) 15 + (D" + -+ (S) "]= X> ^ = [^] ^ = ^, where 
f(x) = x 15 on [0, 1] (see Exercise 25) 

(c) lim - [sin - + sin — + ...+ sin — 1 = I sin mi dx = [— - cos 7rxl . = — - cos 7r — (— - cos 0) 
= -, where f(x) = sin n\ on [0, 1] (see Exercise 25) 

(d) lim 4, [1 15 + 2 15 + ... +n 15 l = ( lim A ( lim -^ \l 15 + 2 15 + . . . + n 15 }) = ( lim ±) f x 15 dx 

= (yg) =0 (see part (b) above) 

(e) lim 4i [1 15 + 2 15 + ... +n 15 l = lim -& fl 15 + 2 15 + .. . + n 15 l 

v ' n — > oo n lj L J n _> oo n 16 L J 

= (^iPoo n ) U^F 1 Sie I 1 " + 215 + ■■■ +n 15 ]) = LliPlo n ) J x 15 dx = oo (see part (b) above) 



27. (a) Let the polygon be inscribed in a circle of radius r. If we draw a radius from the center of the circle (and 
the polygon) to each vertex of the polygon, we have n isosceles triangles formed (the equal sides are equal 
to r, the radius of the circle) and a vertex angle of 8„ where 9 a = — . The area of each triangle is 
A n = i r 2 sin 6 n =4> the area of the polygon is A = nA„ = ^- sin 6„ 



t&in^. 



(b) lim A = lim 



™f s i n 2* = li m 2^ sin 27T _ i; m („ r 2\ si "(v) _ ,..,.:', ,;,„ v„ : r i ^ 



lim (m 2 ) ^^ = (ttt 2 ) lim 



27r/n -> 



28. y = sinx+ I cos2tdt + 1 = sinx — I cos 2t dt + 1 =4> y' = cos x — cos(2x); when x = tt we have 

y' = cos7r — cos(27r) = — 1 — 1 = — 2. And y" = — sinx + 2sin(2x); when x = 7r, y = sin7r + I cos2tdt + 1 



+ 0+1 = 1. 



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358 Chapter 5 Integration 



dt = 



29. (a) g(l) = I f(t 

(b) g(3) = / 3 f(t)dt=-i(2)(l) = -l 

(C) g(-l) = / 'f(t)dt=-j\(t)dt: 



dt = 
0=>x 



>2 2 ) = -tt 



-3, 1, 3 and the sign chart for g'(x) = f(x) is 



-+ | | +++. So g has a 

1 3 



dt 



(d) g'(x)=f(x) 
relative maximum at x = 1 . 

(e) g'(-l) =f(-l) =2 is the slope and g(-l) = J f(t) 
y = 2x + 2 - 7T. 

(f) g"(x)=f'(x)=0atx 

inflection point for g at x = — 1. We notice that g"(x) = f'(x) < for x on (— 1, 2) and g"(x) = f'(x) > for x on 
(2, 4), even though g"(2) does not exist, g has a tangent line at x = 2, so there is an inflection point at x = 2. 

(g) g is continuous on [—3, 4] and so it attains its absolute maximum and minimum values on this interval. We saw in (d) 
that g'(x) = => x = -3, 1, 3. We have that 



-71", by (c). Thus the equation isy + 7r = 2(x+l) 
-1 and g"(x) = f'(x) is negative on (—3, —1) and positive on (— 1, 1) so there is an 



g(-3) = f~ 'fCt 
g(l) = J>( 

g( 4 ) = Ij( 



dt= - 

l ) dt = 



,-(t)dt=-^ 2 



-2tt 



t) dt 



■1 



t dt: 



■1-1 



Thus, the absolute minimum is — 2ir and the absolute maximum is 0. Thus, the range is [— 2ir, 0]. 



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Chapter 5 Additional and Advanced Exercises 359 



NOTES: 



Copyright (c) 2006 Pearson Education, 




360 Chapter 5 Integration 
NOTES: 



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CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS 



6.1 VOLUMES BY SLICING AND ROTATION ABOUT AN AXIS 



1. (a) A = 7r(radius) 2 and radius = y 1 - x 2 =>- A(x) = tt (1 — x 2 ) 

(b) A = width • height, width = height = 2\/l - x 2 =>■ A(x) =4(1 - x 2 ) 

(c) A = (side) 2 and diagonal = y^side) =>- A = (diag ° nal)2 ; diagonal = 2\/l - x 2 =>• A(x) = 2 (1 - x 2 

(d) A = ^ (side) 2 and side = 2\/l - x 2 => A(x) = a/3 (1 - x 2 ) 

2. (a) A = 7r(radius) 2 and radius = -y/x => A(x) = 7rx 

(b) A = width - height, width = height = 2^/x => A(x) = 4x 

(c) A = (side) 2 and diagonal = y^Cside) =4> A = (diag ° nal)2 ; diagonal = 2^/x =>■ A(x) = 2x 

(d) A = ^ (side) 2 and side = 2\/x =>• A(x) = a/^x 

3. A(x) = (dlas ° nal)2 = (^-(-\A)) 2 = 2x (see Exercise lc); a = 0, b = 4; 
V = f A(x) dx = J 2x dx = [x 2 ] * = 16 



4. A (x) = " (dla 7 ter)2 = " 1(2 - x p- x2]2 = " [2(1 - x2) ' 2 = tt (1 - 2x 2 + x 4 ) ; a = -1, b = 1; 



V = J] A(x) dx = J_ t tt (1 - 2x 2 + x 4 ) dx = tt I" 



x-|x3 + ^^=2.(l-| + i) = if 



5. A(x) = (edge) 2 = \yf\ - x 2 - (-\/l - x 2 )] = (ly/i - x 2 V =4(1 - x 2 ) ; a = -l,b= 1; 
V = £ A(x) dx = f_ t 4(1 - x 2 ) dx = 4 |x - f 1 =8(1- i) 



16 
3 



6. A(x) 



(diagonal)- 



al) 2 _ [fi^-{-vv^)\ 2 _ (is/r^y 



2(1— x 2 ) (see Exercise lc); a = — 1, b = 1; 



V = J A(x) dx = 2J_ t (1 - x 2 ) dx = 2 [ 



*■ T|_ 1 =4(l-I) = | 



7. (a) STEP 1) A(x) = \ (side) • (side) - (sin |) = \ ■ (l\Jsmx\ ■ [l\/sm x\ (sin |) = \fb sin x 

STEP 2) a = 0, b = tt 

STEP 3) V = J a A(x) dx = v/3 JJsin x dx = V-y/l cos xj " = \fl{\ + 1) = 2^/3 

(b) STEP 1) A(x) = (side) 2 = (ly/sin x\ (ly/sm x\ = 4 sin x 

STEP 2) a = 0, b = tt 

STEP 3) V = J A(x) dx = f 4 sin x dx = [-4 cos x] * = 8 



8. (a) STEP1) A(x) = ^^£_£ = | ( S ec x - tan x) 2 = f (sec 2 x + tan 2 x - 2 sec x tan x) 



[sec 2 x+(sec 2 x-l)-2^] 



4 



STEP 2) a=-f,b 



3' " ~~ 3 

b rirft 



STEP 3) V = J a A(x) dx = f | (2 sec 2 x - 1 - ^f ) dx = f [2 tan x - x + 2 (- ^ 



-)V /3 

x/J-tt/3 



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362 Chapter 6 Applications of Definite Integrals 

= l[2 V ^-! + 2(- I } y )-(-2y3 + f+ 2(- I j y ))]=|(4y3^) 

(b) STEP 1) A(x) = (edge) 2 = (sec x - tan x) 2 = (2 sec 2 x - 1 - 2 |j£) 
STEP 2) a= -f,b= f 

STEP 3) V = £A(x) dx = £^ (2 sec 2 x - 1 - f$i) dx = 2 (2^3 - f ) = 4 ^3 - f 



9. A(y) = | (diameter) 2 = | (\/5y 2 - o) = & y 4 ; 
c = 0, d = 2; V = / c A(y) dy = £ 5 f y 4 dy 

= [(f) (f ) 



E M5 



(2 5 - 0) = 8tt 



1.5 



0.5 




o+ — j — r~ s 



10. A(y) = i (leg)(leg) = 1 [^l^f - (-/T^y 2 )] 2 = \ {l^T^ff = 2 (1 -y 2 ) ; c = -1, d = 1; 

V = / c d A(y)dy = /_ 1 i 2(l-y 2 )dy = 2[y-^^=4(l-I) = | 

11. (a) It follows from Cavalieri's Principle that the volume of a column is the same as the volume of a right 

prism with a square base of side length s and altitude h. Thus, STEP 1) A(x) = (side length) 2 = s 2 ; 

STEP 2) a = 0, b = h; STEP 3) V = J A(x) dx = J q s 2 dx = s 2 h 
(b) From Cavalieri's Principle we conclude that the volume of the column is the same as the volume of the 
prism described above, regardless of the number of turns =>• V = s 2 h 



12. 1) The solid and the cone have the same altitude of 12. 

2) The cross sections of the solid are disks of diameter 
x— (|) = |. If we place the vertex of the cone at the 
origin of the coordinate system and make its axis of 
symmetry coincide with the x-axis then the cone's cross 
sections will be circular disks of diameter 

| — (— |) = | (see accompanying figure). 

3) The solid and the cone have equal altitudes and identical 
parallel cross sections. From Cavalieri's Principle we 
conclude that the solid and the cone have the same 
volume. 




13. R(x) =y=l-|=>V= XV[R(x)] 2 dx = n£(l - |) 2 dx = tt/^I - x + f ) dx = tt [x - f + £ 



7T(2- 



2 ' UJ 3 



14. R(y) = x = | =► V = £ ^[R(y)] 2 dy = ^£{^f dy = tt £\ y 2 dy = tt [| 



,31 I = n . 3 . g = ^ 



15. R(x) = tan (| y) ; u = | y => du = | dy => 4 du = tt dy; y = => u = 0, y = 1 => u = f ; 

V = f o 7r[R(y)] 2 dy = n£ [tan (f y)] 2 dy = 4 £ "tan 2 u du = 4 £ (-1 + sec 2 u) du = 4[-u + tan u]q /4 

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Section 6.1 Volumes by Slicing and Rotation About an Axis 363 



4(-f + l-0)=4-7r 



C T / 2 r, 

16. R(x) = sin x cos x; R(x) = => a = and b = | are the limits of integration; V = I 7r[R(x)] dx 



Trf* (sin x cos x) 2 dx = tt J* (si " 4 2x)2 dx; [u = 2x => du = 2 dx =4> 



| = ^;X = => U = 0, 



u = 7 r] - V = ^/;isin 2 udu=|[f-isin2u]; = f [(f-0)-0] 



17. R(x) = x 2 =4> V = f~ ?r[R(x)] 2 dx = tt / ~(x 2 ) 2 dx 



7 r/Vdx = 7 rff 



32tt 

5 




18. R(x) = x 3 =4> V = J tt[R(x)] 2 dx = wj (x 3 ) 2 dx 



1287T 

7 




19. R(x) = ^9-x 2 =4> V = J_ 3 7r[R(x)] 2 dx = tt f (9 - x 2 ) dx 



tt 9x 



2tt [9(3) - f ] = 2 • tt • 18 = 36tt 



20. R(x) = x - x 2 =>• V = J o tt[R(x)] 2 dx = ttJ o (x - x 2 ) 2 dx 
= tt£{v? - 2x 3 + x 4 ) dx = TT U 



2x1 J_ J! 
3 4 + 5 



.3 2 ' 5^ 30 



^(10- 15 + 6)= ^ 



30 




'■tt/ 2 



21. R(x) = ^/cosx =>• V = J* 7r[R(x)] 2 dx = tt J* cos x 



dx 



tt [sin x] q = 7r(l — 0) = 7T 



y=(cosx) 1/2 




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364 Chapter 6 Applications of Definite Integrals 



22. 



/tt/4 /»7t/4 

7r[R(x)] 2 dx = 7t / sec 2 x 
-7r/4 J —tt/4 



dx 



7r[tanx]!^ /4 = 7r[l-(-l)]=27r 




1 
0.8 
0.6 



23. R(x) = \fl - sec x tan x => V = £ tt[R(x)] 2 dx 

fV 4 / i- \ 2 

= 7r / v2 — sec x tan x dx 
Jo \ v / 

= 7r I ( 2 — 2 v 2 sec x tan x + sec 2 x tan 2 x J dx 

( r n/4 r p 7 ^ 4 n /4 ■? ■> \ 

= 7r I / 2dx — 2\/2j sec x tan x dx + I (tan x) sec x dx I <>•'• 

\ / 0.2 

= n f[2x]^ /4 - 2 V / 2[sec x]f + [^] ^ 

= ^[(|-0)-2y2(y2-l)+i(l 3 -0)]=7r(f + 2y2-^) 



y = V2 




y = sec i tan x 



24. R(x) = 2 - 2 sin x = 2(1 - sin x) =>■ V = F ~tt[R(x)] 2 dx 

** 

X7r/2 f*V 2 

4(1 — sin x) 2 dx = 47r I (1 + sin 2 x — 2 sin x) dx 

= 47rJ" r [l + 1 (1 - cos 2x) - 2 sin x] dx 



^L ( 



4tt 4 x - 



| _ co|2x _ 2 sin x j 



i2x 



+ 2 cos x 



tt/2 



L2 "■ 4 ' - — ""JO 

= 4tt [(f - + 0) - (0 - + 2)] = ttOtt - 8) 

25. R(y) = v/5 • y 2 =* V = £ ^[R(y)] 2 dy = it £ 5y 4 dy 
= 7T [y 5 ] ^ = tt[1 - (-1)] = 2tt 




1 x 




*■ X 



26. R(y) = y 3 / 2 => V = J^ ^[R(y)] 2 dy = ^y 3 dy 



47T 




27. R(y) = V2 sin 2y =* V = £ ^[R(y)] 2 dy 

r /2 r n - 

= 7r / 2 sin 2y dy = ir [— cos 2yJ j 
= tt[1 -(-1)] = 2tt 



i;' 2 



Copyright (c) 2006 Pearson Education 




Section 6.1 Volumes by Slicing and Rotation About an Axis 

28. R(y) = ^/cosf => V = J%[R(y)] 2 dy y 

= 7T /° 2 COS (f ) dy = 4 [sin f ] °_ 2 = 4[0 - (-1)] = 4 



365 



29. R(y) = ^ =* V = |>[R(y)] 2 dy = 4* £ ^ dy 



47T 



y+i 



47r[-i -(-!)] =37 





* = 2/(»+l) 



30. R(y) = ^ => V = J o ' 7r[R(y)] 2 dy = ir£ 2y (y 2 + 1)~ 2 dy; 
[u = y 2 + 1 =^> du = 2y dy; y = =4> u = 1, y = 1 => u = 2] 

- v = tt £ u- 2 du = tt [- 1] ; = n [- 1 - (-D] = i 



x = V^/(s/ , + l) 




TJ3 5!3 53 BT" 



31. For the sketch given, a = - § , b = § ; R(x) = 1, r(x) = v /cosx; V = J] tt ([R(x)] 2 - [r(x)] 2 ) dx 

= 1 7r(l — cos x) dx = 27r I (1 — cos x) dx = 27r[x — sin x]q = 27r (| — l) = 7r 2 — 2tt 

32. For the sketch given, c = 0, d = f ; R(y) = 1, r(y) = tan y; V = J tt ([R(y)] 2 - [r(y)] 2 ) dy 

= tt£'\i - tan 2 y) dy = tt £'\l - sec 2 y) dy = 7r[2y - tan y]^ 4 =7r(f-l) = ^-7T 

33. r(x) = x and R(x) = 1 =>- V = / 7T ([R(x)] 2 - [r(x)] 2 ) dx 



XV(l-x 2 )dx = .[x-f] o =^[(l-I)-0] 



277 

3 




34. r(x) = 2^/x and R(x) = 2 =>■ V = J 7r ([R(x)] 2 - [r(x)] 2 ) dx 
= 7 r/ o 1 (4-4x)dx = 4 7 r[x-f]^4 7 r(l-i)=27r 




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366 Chapter 6 Applications of Definite Integrals 



35. r(x) = x 2 + 1 and R(x) = x + 3 

=> V = J%([R(x)] 2 -[r(x)] 2 )dx 

= 7r£ [(x + 3) 2 -(x 2 + l) 2 ]dx 

= 7T f [(x 2 + 6x + 9) - (x 4 + 2x 2 + 1)] dx 

= 7T f (-x 4 - x 2 + 6x + 8) dx 




i x 



., _ £ _ £ + 6x! + g x 



7r [(-f-| + M + 16) _ ( i + I + |_ 8)]=7r( _ f _ 3 + 28-3 



_ 1 5-3 0-33 \ Wir 



36. r(x) = 2 - x and R(x) = 4 - x 2 

=> V = /_", ^([R(x)] 2 -[r(x)] 2 )dx 

= *£ [( 4 - x2 ) 2 - ( 2 - x > 2 ] dx 

= 7T /_* [(16 - 8x 2 + x 4 ) - (4 - 4x+ x 2 )] dx 

= n J (12 + 4x - 9x 2 + x 4 ) dx 



7T 1 12x + 2x 2 - 3x 3 + f 



= 7T [(24 + 8 - 24+ f ) - (-12 + 2 + 3 - i)] = 7T (15 
37. r(x) = sec x and R(x) = y 2 

(2 — sec 2 x) dx = 7r[2x — tan x]_ , 4 
= 7r[(f-l)-(-f + l)]=7r(7r-2) 




33 \ _ 108tt 



y 




y = secx 


Ht/4 


*/4 



38. R(x) = sec x and r(x) = tan x 

=> V = J%([R(x)] 2 -[r(x)] 2 )dx 

= 7r I (sec 2 x — tan 2 x) dx = 7r I 1 dx = 7r[x]J = 7r 




39. r (y) = 1 and R(y) = 1 + y 

=> V = J%([R(y)] 2 -[r(y)] 2 )dy 

= ir£ [(1 + y) 2 - 1] dy = 7T £ (1 + 2y + y 2 - 1) dy 

= .X 1 (2y + y 2 )dy = 7 r[y 2 + ^;=.(l + I) = 4 



y 




=x-1 



< X 



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Section 6.1 Volumes by Slicing and Rotation About an Axis 367 



40. R(y) = 1 and r(y) = 1 - y =>• V = f tt ([R(y)] 2 - [r(y)] 2 ) dy 

*J 

= irf o [1 - (1 - y) 2 ] dy = tt £ [1 - (1 - 2y + y 2 )] dy 

= n I ( 2 y ~ y 2 ) d y = ^ [> 



\y 2 - y i 



TT 1 



n _ 2tt 




41. R(y) = 2andr(y) = ^y 

=> V = f Q 7T ([R(y)] 2 - [r(y)] 2 ) dy 

= ttJ q (4 - y) dy = tt Uy - y] = tt(16 - 8) = 8tt 




42. R(y) = v/3 and r(y) = ^3 - y 2 

=* V = / o 3 tt ([R(y)] 2 - [r(y)] 2 ) dy 

= it jf [3 - (3 - y 2 )] dy = tt ^ y 2 dy 



;rl T 



ttV^ 




t^x 



43. R(y) = 2 and r(y) = 1 + ^/y 

=> V = / o 1 7r([R(y)] 2 -[r(y)] 2 )dy 

= -r[ 4 -0 + ^) 2 ] d y 

= 7r / o 1 (4-l-2^/y-y)dy 

= ^X'( 3 - 2 ^-y) d y 



4 3/2 _ r 



2 Jo 




M 6 J = 


7?r 
" 6 



tt |3y - f y 
- (3 - f " | 



44. R(y) = 2 - y 1 / 3 and r(y) = 1 

=> V = J%([R(y)] 2 -[r(y)] 2 )dy 

= -X l [( 2 -y 1/3 ) 2 - 1 ] d y 

= t:£ (4 - 4y 1 / 3 + y 2 / 3 - l) dy 

= tt f (3 - 4y 1 < /3 + y 2 / 3 ) dy 

= tt [3y - 3y 4 / 3 + ^1 * = tt (3 - 3 + 



3^r 
5 




c 




y 



=-x3 



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368 Chapter 6 Applications of Definite Integrals 



45. (a) r(x) = ^/x and R(x) = 2 

=>■ V = / ) 4 7r([R(x)] 2 -[r(x)] 2 )dx 
= 7T J u (4 - x) dx = 7T Ux - f 1 = 7T(16 - 8) = 8tt 
(b) r(y) = and R(y) = y 2 

=> V = £7r([R(y)] 2 -[r(y)] 2 )dy 

r 2 

= 7T J o y 4 dy = 7T 



32g 

5 




(c) r(x) = and R(x) = 2 - yfi. => V = £ it ([R(x)] 2 - [r(x)] 2 ) dx = tt£ (2 - ^/x") 2 dx 



7rJ (4-4-y/x + x) dx = 7r 4x- 



8x 3 / 2 i x 2 
3 " l " 2 



7T(16 



64 _, 16^1 8tt 

3 "r" 2 ) ~ 3 



1 X 



(d) r(y) = 4 - y 2 and R(y) = 4 => V = £ tt ([R(y)] 2 - [r(y)] 2 ) dy = tt J^ [l6 - (4 - y- f | dy 

2 



tt / o (16 - 16 + 8y 2 - y 4 ) dy = it J q (8y 2 - y 4 ) dy = tt [f y 



3 _ r 

5 



'64 32 \ _ 224?r 



46. (a) r(y) = and R(y) = 1 - § 

=*• v = X, 2 * ( [R( y )]2 - [r( y )]2 ) d y 
= -X 2 ( 1 -D 2d y = -r( 1 -y + T)dy 

(b) r(y) = 1 and R(y) = 2 - \ 




- V = £ tt ([R(y)] 2 - [r(y)] 2 ) dy = tt £ [(2 - \f - l] dy = tt £ (a - 2y + J - l) dy 

-r(3-2y+£)dy = 7 r[3y-y 2 + 1 C=.(6-4+A) =7r (2 + !) 



8tt 
37 — 3 



47. (a) r(x) = and R(x) = 1 - x 2 

=* V = /_'^([R(x)] 2 -[r(x)] 2 )dx 

= tt J_* (1 - x 2 ) 2 dx = tt £ (1 - 2x 2 + x 4 ) dx 

= 7r[x-f + ^] ^ = 27T (1 - | + I) 



2,(15^3) = m 




1 

(b) r(x) = 1 and R(x) = 2 - x 2 =J> V = £ tt ([R(x)] 2 - [r(x)] 2 ) dx = tt £ \(2 - x 2 ) 2 - lj dx 

= tt £ (4 - 4x 2 + x 4 - 1) dx = tt£ (3 - 4x 2 + x 4 ) dx = tt hx - f x 3 + f 1 = 2tt (3 - f + ±) 



ff (45 - 20 + 3) = Sgr 



(c) r(x) = 1 + x 2 and R(x) = 2 =>■ V = J*_ [ tt ([R(x)] 2 - [r(x)] 2 ) dx = tt £ U - (1+ x 2 ) 2 ] dx 
= tt /_' (4 - 1 - 2x 2 - x 4 ) dx = tt£ (3 - 2x 2 - x 4 ) dx = tt hx - § x 3 - f 1 = 2tt (3 - 



2 _ I N 

3 5/ 



£(45-10-3)= Sfe 



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Section 6.1 Volumes by Slicing and Rotation About an Axis 369 



48. (a) r(x) = and R(x) = - j) x + h 

=> V = £7r([R(x)] 2 -[r(x)] 2 )dx 

=^r(-s x+h ) 2dx 



*£( 



f 2 x 2 - ^ x + h 2 ) <lx 



;.-lr 1^- y+x|" - -..lr f 1 : b + b) 



^h 2 b 
3 



=-(h/b)x+h 




(b) r(y) = OandR(y) = b(l-£) =>■ V = ]% ([R(y)] 2 - [r(y)] 2 ) dy = Trb 2 /^ (l - I) 2 dy 



49. R(y) = b + ,/a 2 - y 2 and r(y) = b - ^/a 2 - y 2 
=>■ V=/_ a a 7r([R(y)] 2 -[r(y)] 2 )dy 
= * /_", [( b + V^f) 2 - (b - y^^7) 2 ] dy 
= 7T J_ 4b^/a 2 - y 2 dy = 4b7rJ_ i/a 2 - y 2 dy 
= 4b7r • area of semicircle of radius a = 4b7r • ^f- = 2a 2 b7r 2 




50. (a) A cross section has radius r = ^/2y and area 7rr 2 = 27ry. The volume is I 27rydy = tt [y 2 ] = 257T. 

l dy 

A(h) ' dt ' 



(b) V(h) = /A(h)dh, so f = A(h). Therefore f = f • f = A(h) ■ f , so f - ! dV 



For h = 4, the area is 2tt(4) = 8?r, so f = gL • 3 



i^ _3_ units 3 

dt 87r " sec 8tt sec 



51. (a) R(y) = v'a 2 - y 2 => V = tt J " (a 2 - y 2 ) dy = tt |"a 2 y - 

= tt [a 2 h - i (h 3 - 3h 2 a + 3ha 2 - a 3 ) - f 1 = tt (a 2 h - f + h 1 



^| a 2 h _ a 3_(h^a£_ (_ a 3 



01 



a — ha' 



-h 2 (3a-h) 
3 



(b) Given ^ = 0.2 m 3 /sec and a = 5 m, find § | h=4 . From part (a), V(h) 



=* ^ = 107rh-rf => ^ = dv.^=^h(10-h)^ =* 



dt ~~ dh dt 



7rh 2 (15-h) 
3 
0.2 



57Th 

1 



2 ?rh 3 



dh I 

dt I h=4 4tt(10-4) (20tt)(6) 120tt 



k m/sec. 



52. Suppose the solid is produced by revolving y = 2 — x about 
the y-axis. Cast a shadow of the solid on a plane parallel to 
the xy-plane. 
Use an approximation such as the Trapezoid Rule, to 



P 2 

estimate / 7r[R(y)] dy « J^7r 



k-l 



Ay. 




53. The cross section of a solid right circular cylinder with a cone removed is a disk with radius R from which a disk of radius 
h has been removed. Thus its area is Ai = 7rR 2 — 7rh 2 = tt (R 2 — h 2 ) . The cross section of the hemisphere is a disk of 

radius \/R 2 — h 2 . Therefore its area is A-2 = tt ( y R 2 — h 2 j = tt (R 2 — h 2 ) . We can see that Ai = A2. The altitudes of 

both solids are R. Applying Cavalieri's Principle we find 

Volume of Hemisphere = (Volume of Cylinder) - (Volume of Cone) = (ttR 2 ) R - \tt (R 2 ) R = § ttR 3 . 

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370 Chapter 6 Applications of Definite Integrals 
54. R(x) = £ => V = XV[R(x)] 2 dx = tt/^ dx = f [l 



h2 



) ( T ) = 5 7rI ' 2 h, the volume of 



a cone of radius r and height h. 



55. R(y) = V256 - y 2 =► V = / ^[RCy)] 2 dy = tt/ J256 - y 2 ) dy = n [256y - £ 

= 7T [(256)(-7) + f - ((256)(-16) + 1 f)] = * (y + 256(16 - 7) - ^) = 1053tt cm 3 « 3308 cm 3 

56. R(x) = fr 2 \/36 - x 2 => V = £ tt[R(x)] 2 dx = irf* £ (36 - x 2 ) dx = fo £ (36x 2 - x 4 ) dx 



12 

12x 3 



144 ^ 



12 - 6 



3 _ £ 

5 



y 144 ^ 



12 



f) 



196tt \ / 60-36 1 



367i 



cm 3 . The plumb bob will 



weigh about W = (8.5)(^p) « 192 gm, to the nearest gram. 



57. (a) R(x) = |c — sin x| , so V = 7r I [R(x)] 2 dx = it I (c — sin x) 2 dx = tt I (c 2 — 2c sin x + sin 2 x) dx 



tt/* (c 2 - 2c sin x + 1 ~ c ° s2x ) dx = tt£ (c 2 + | - 2c sin x - 



cos 2x\ 



dx 



= tt [(c 2 + \) x + 2c cos x - Siii] = tt [(c 2 tt + | - 2c - 0) - (0 + 2c - 0)] = tt (c 2 tt + | - 4c) . Let 
V(c) = n (c 2 7T + I — 4c) . We find the extreme values of V(c): ^ = 7r(2c7r — 4) = => c = -^ is a critical 
point, and V (f ) = tt (£ + f - f ) = tt (f - £) = ^ - 4; Evaluate V at the endpoints: V(0) = y and 
V(l) = 7r (| 7r — 4) = \ — (4 — 7r)7r. Now we see that the function's absolute minimum value is ^ — 4, 
taken on at the critical point c = -. (See also the accompanying graph.) 

2 

(b) From the discussion in part (a) we conclude that the function's absolute maximum value is y , taken on at 
the endpoint c = 0. 

(c) The graph of the solid's volume as a function of c for 
< c < 1 is given at the right. As c moves away from 
[0, 1] the volume of the solid increases without bound. 
If we approximate the solid as a set of solid disks, we 
can see that the radius of a typical disk increases without 
bounds as c moves away from [0, 1]. 




58. (a) R(x) = 1 - f| =► V = /_ 4 4 7r[R(x)] 2 dx 

= -[*-i + 5^] 4 _ 4 = 2 -( 4 -! + 5^) 



2tt(4 



! + !) 



24 ' 5-1 

(60 - 40 + 12) 



64?r ft 3 
15 ll 



1-(x2/16) 




(b) The helicopter will be able to fly (^f ) (7.481)(2) « 201 additional miles 



6.2 VOLUME BY CYLINDRICAL SHELLS 



1. For the sketch given, a = 0, b = 2; 

V = J> (X s ) (hth.) d X = />X (l + *) dx = 27T/; (X + £) dx = 27T [i + i] ' = 27T 

= 27T • 3 = 67T 



2 T lb) 



2. For the sketch given, a = 0, b = 2; 

V = J> (** ) (*J) dx = />rx (2 - f ) dx = 27r/; ( 2 x - f ) dx = 2tt [x 



2 £ 

16 



2tt(4 - 1) = 6tt 



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Section 6.2 Volume by Cylindrical Shells 371 



3. For the sketch given, c = 0, d = y2; 



v = r^(^)(^ t ) d y = X V "^y(y 2 )dy = 2^y 3 dy = 2.p]f = 2. 



4. For the sketch given, c = 0, d = y 3; 



•v^> 



v^ 



v = X 2. ( ** ) (St) d y = /. 2 -y ■ p - ( 3 - y 2 )] d y = 2 - X ' y 3 d y = 2 - [J 

5. For the sketch given, a = 0, b = y 3; 

v = J> (X.) (it) d * = J^™ • (v 7 ^) dx; 

| u = x 2 + 1 =4> du = 2x dx; x = =4> u = 1, x = \/3 => u = 4 1 

-> V = ^ u 1/2 du = 7T [| u 3 / 2 ] J = f (4 3 / 2 - 1) = (f ) (8 - 1) 



41 v/3 



9^ 
2 



14- 
3 



6. For the sketch given, a = 0, b = 3; 

— J a Vradius/ ^ height^ ~~ J \^/y? + 9 ) ' 

[u = x 3 + 9 =>• du = 3x 2 dx => 3 du = 9x 2 dx; x = : 
-^ V = 2tt /"3U- 1 / 2 du = 6. [2u J / 2 ] ^ 6 = 12tt (y^6 



J,x = 3 

= 36tt 



u = 36] 



7. a = 0, b = 2; 



V = J> (** ) (**) dx = X 2 2.x [x -(-!)] dx 
= X, 2 2ttx 2 - 1 dx = 7T X 3x 2 dx = 7T [x 3 ] \ = 8tt 




a = 0, b= 1; 

V = j> GEL) (S) ^ = X, 1 2 ™ ( 2x " 1) dx 

= .X' 2 (f)dx = .i3x 2 dx = 7 r[x 3 ]; = . 




9. a = 0,b= 1; 



V = X> (££) (£t) dx = X' 2 ™ K2 - x) - x 2 ] dx 
= 2ttX (2x - x 2 - x 3 ) dx = 2tt [x 2 - f - 



2^(1-1-^ 



2tt 



' 12-4-3'" 



IOtt _ Stt 
12 — 6 



y=2-x 




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372 Chapter 6 Applications of Definite Integrals 



10. a = 0,b= 1 



V = J> ( ** ) (**) dx = / o 2.x [(2 - x*) - x 2 ] dx 

= 2tt £ x (2 - 2x 2 ) dx = 4tt/ o (x - x 3 ) dx 

= 47T 



Ml-i) 



11. a = 0,b= 1; 

V = j> (££) (S) ^ = /.' 2 ™ [^ - (2X - 1} ] ^ 

= 2ttJ* (x 3 / 2 - 2x 2 + x) dx = 2tt [§ x 5 / 2 



3 X + 2 X J Q 



2ir 



3 ' 2 J 



2tt 



' 12-20+ 15 1 

30 ; 



7rr 
15 



12. a= l,b = 4; 

V = J> ( r S) (it) dx = J> x (I x " /2 ) dx 
= 3tt J^x 1 / 2 dx = 3ir [f x 3 / 2 ] \ = 2ir (4 3 / 2 - l) 
= 2tt(8 - 1) = 14tt 





2 

1.5 + 

1 
0.5 



tf5 !TS 1 



2Vx" 




13. (a) xf(x) = <^ => xf(x) = <^ 

^ x, x = ^ 



sin x, < x < 7r 
0, x = 



since sin = we have 



xf(x) = | 



sin x, < x < 7r 
sin x, x = 



=>- xf(x) = sin x, < x < 7r 



(b) V = f\w (*£) ( *gj dx = Io 2lTX ■ f(x) dx and x • f(x) = sin x, < x < 7T by part (a) 

=>• V = 2ir J sin x dx = 2ix\— cos x]J = 27r(— cos 7r + cos 0) = 47r 



14. (a) xg(x) 



<x<| 
x = 

,2 v n < x < tt/4 

x = 



xg(x) 



{ 



tan 2 x, < x < 7i74 



0, x = 



since tan = we have 



1 x-0, x 

f tan 2 x, 
xg(x) = i . 2 

[ tan~ x, x 

(b) V = fl-K (*£) (*St) dx = JJ^ttx • g(x) dx and x • g(x) = tan 2 x, < x < tt/4 by part (a) 
=> V = 2tt J 7 " tan 2 x dx = 2?r J (sec 2 x - 1) dx = 2?r[tan x - x][/ 4 = 2tt (l 



xg(x) = tan 2 x, < x < 7r/4 

•>tt/4 



At; — 7t 2 



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15. c = 0,d = 2; 



v = J> (SJ (S) d y = J>y [^ - (-y)] d y 

=2tt r 

Jo 



y 2 ) dy = 2?r 



v y . j y ^ - - s 

5 



^(3^ + 5) 



2tt 



(?# + f) = 16.(f + i) 



Section 6.2 Volume by Cylindrical Shells 

y 



373 





2 




x=-y ^ljp 


#x => /y 


-2 




vk 



16. c = 0, d = 2; 

v = j> GEL) (S) ^ = T 2 -y [y 2 - (-y)]dy 



2^(y 3 + y 2 ) d y = 2 ^fT+T 



167r(| + i) 



\6tt 1 



4()tt 




t— X 



17. c = 0, d = 2; 

v = J> ( r S) (St) d y = 1 2 -y ( 2 y - y 2 ) d y 

= 2^;(2y 2 - y 3)dy = 2 7 r[¥-fl' = 2 -(f^) 



32tt « 



12 3 



x=2y-y : 




1 x 



18. c = 0, d= 1; 

v = J> (£.) (S) d y = /.' 2 -y ( 2 y - y 2 - y) d y 

= 2tt£ y (y - y 2 ) dy = 2n£ (y 2 - y 3 ) dy 



2n\\-\ 



^{\-\) 




19. c = 0,d= 1; 



v = J> (SJ (S) d y = 2 -X' y[y - (-yw 

= 2^ , 2y 2 dy=f [ y 3]J 



47T 

10 3 




20. c = 0, d = 2; 



v = r 2 -( r ^)(S) d y = r 2 -y(y-D d y 

= 2 -r^y=f[y 3 ] 



1 _ 8tt 
— 3 




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374 Chapter 6 Applications of Definite Integrals 



21. c = 0, d 

ml 



V = I 2- ( ** ) (**) dy = / o 2.y [(2 + y) - y 2 ] dy 

= 2n £ (2y + y 2 - y 3 ) dy = 2. [y 2 + £ - 



2tt 4 



16 



±2 = | (48 + 32 - 48) 



16- 
3 




22. c = 0, d = 1 



V = I 2n (** ) (**) dy = X 27 ry [(2 - y) - y 2 ] dy 

= 2n£ (2y - y 2 - y 3 ) dy = 2. [y 2 - f - 



Mi-l-i) 



(12-4-3) 



6 




23. (a) V = X 2. ( 



shell \ / shell 
radius / \ height 



dy = X' 2ny - 12 (y 2 - y 3 ) dy = 24. £ (y 3 - y 4 ) dy = 24. [£ - f ] * 



24. 



' i r 



24_7T _ 67T 

20 — 5 



w v = X> (Xs) (h s S0 d y = 1 2 ^ - y) t 12 (y 2 - y 3 )] d y = 24 -X' o - y) (y 2 - y 3 ) d y 

= 24.X 1 (y 2 -2y 3 + y 4 )dy = 24 7 rp-^ + ^; = 24.(1-1 + I) =24.(3^) = f 

(o v = X> (It) (fit) d y = X' 2 - (! - y) I 12 (y 2 - y 3 )i d y = 24 - /.' (I - y) (y 2 - y 3 ) d y 



24 ^X 1 (sy 2 -Ty 3 + y 4 ) d y = 247r [ 



8 „3 13 „4 



20 



y 4 + 



24.! 



13 , n 

20 T 5 y 



2-1;; 
60 



(32 - 39 + 12) 



2. 



(d) V = ./>(**) ( h ^ t ) dy = X' 2^ (y + I) [12 (y 2 - y 3 )] dy = 24n£ (y + §) (y 2 - y 3 ) dy 

= 24 -X 



(y 3 - y 4 + I y 2 - I y 3 ] 



d y = 24 ^X (I y 2 + 1 y 3 - y 4 ) d y = 24. | ^ >■■■ + :n >< ■ 



2 „3 1 3 „4 



2^(^ + h 5) = W (8 + 9 - 12) = 2 t = 2. 



24. (a) V = X 2* (*£ ) ( -J) dy = X 2.y f? - (J - £)] dy = J~ 2.y (y 2 - $) dy = 2^ (y 3 - £) dy 
= 2 7 r[^-g] 2 o =2 7 r($-|) = 32* (J - £) = 32* (J - $) = 32* (£) = f 
(b) V = X> (*£) (** ) dy = X>(2 - y) [£ - ($ - £ )] dy = £ 2.(2 - y) (y 2 - £) dy 

= 2 -X 2 ( 2 y 2 -^-y 3 + fldy = 2-[¥-^-^ + fl^ = 2-(f-i-f + |) = f 
(0 v = J> (,££) (5t) d y = 1 2 -( 5 - y) [? - (S - ?)] dy = X^ 5 - y) (y 2 - 1) d y 

= 2 -r(V~!y 4 -y 3 + ?)dy = 2 7 r[f-|f-^ + ^] 2 = 2.(f-if-if + |)=8 7 r 

(d) v = X^( r r)(he; g L)dy = X 22 -(y + ep-p-fl]dy = X 22 -(y + I)(y 2 -^)dy 



2tt f ( v 3 - y - + - v 2 - - v 4 ^l dv = 2. \ y - - ^ + ^f - Ml = 2. (16 _ w + 40 _ 160\ = 4 

Jo V 4 ^ 8 y 32 y J u > ^" [4 24 T 24 160j „ ^ U 24 ^ 24 160^ ^" 



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Section 6.2 Volume by Cylindrical Shells 375 



25. (a) About x-axis: V= J>(*£) (*j)dy 

= I" 2 ^y(yy - y)dy = ^/J (y 3/2 - y 2 ) d y 

= 2-[§y 5/2 -|y 3 ]: = 2.(!-i) = i 

Abouty-axis:V=/>( l S s )(h5t) dx 

= J 2ttx(x - x 2 )dx = 2?r J (x 2 - x 3 )dx 



2tt 



i = 27r (3-l) 




(b) About x-axis: R(x) = x and r(x) = x 2 => V = J tt[R(x) 2 - r(x) 2 ]dx = J tt[x 2 - x 4 ]dx 



.3 5J 



2£ 

1") 



About y-axis: R(y) = ,/y and r(y) = y => V = J -k [R(y) — r(y) ] dy = J 7r[y — y 2 ]dy 



| f _ yf 

" ' 2 3 



,2 3^ 



26. (a) V = J\r[R(x) 2 -r(x) 2 ]dx = 7rJ ) 4 [(f + 2) 2 -x 2 ]dx 

= TrJ^-fx 2 + 2x + 4)dx = 7r[-f + x 2 + 4x1 
= tt(-16 + 16 + 16) = 16tt 
(b) V = /> ( ** ) (**)«* = />rx(§ + 2 - x)dx 

= lW-i)dx = 2^ o 4 (2x-f)dx 




< X 



(c) V = />(*£) (ht".) dx = SM A " x )(f + 2 - *)d* = J>(4 -x)(2 - |)dx = 2^(8 - 4x + % )dx 



2tt 8x - 2x 2 



27r(32-32 + 



64 \ _ 64?r 



(d) V = _{ i b 7r[R(x) 2 -r(x) 2 ]dx = 7r/ o 4 [(8-x) 2 - (6- | ) 2 ]dx = 7r^ 4 [(64 - 16x + x 2 ) - (36-6x+^) 
irf Q (|x 2 -10x + 28)dx = 7r[^ - 5x 2 + 28x1 =tt[16- (5) (16) + (7) (16)] = tt(3)(16) = 48tt 



dx 



27. (a) V = J> (** ) ( -J) dy = J>y(y - I) dy 

= 2^ 2 (y 2 - y) dy = 2, [f - f ] ' 



2tt (I - 2 + i) = | (14 - 12 + 3) = % 



2 




(b) V = J%(**)(**)dx 

= f*2-irx(2 - x) dx = 2ir£(2x - x 2 ) dx = 2tt [x 2 - f 1 = 2tt [(4 - |) - (l - ±)] 

= 2.[(^)-( 3 T l)]=2 7 rG-f) = f 

w v = /> (.s) (it) dx = r 2 - (t - x ) ( 2 - x ) dx = 2 -r 



'20 _ 16 

. 3 3 



x + x 2 ) dx 



2.[fx-fx 2 + Ix3] 2 = 2.[(f-f + f)-(f-| + I)]=2.(|)=2. 



(d) V=/ c d 27T( 



shell \ / shell 
radius / \ height 



dy = f*2v(y - l)(y - 1) dy = 2^/^y - l) 2 = 2tt [ 



(y-D J 
3 



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376 Chapter 6 Applications of Definite Integrals 



28. (a) V = J> ( ** ) (**) dy = filiytf - 0) dy 

= 2.J;y 3 dy = 2. [^ = 2.(^=8. 

(b) V=/>(^j(^ t )dx 

= f*2nx (2 - y/x) dx = 2?r/ o (2x - x 3 / 2 ) dx 
= 2.[x 2 -§x 5 / 2 ] 4 , = 2.(l6- 2 ^) 



2.(16- f,- 5 



2yr (80 - 64) = ^ 




(0 V = J> (5£) (5t) dx = I 2 ^ 4 " x ) ( 2 - V^) ^ = 2./> - 4xV 2 - 2x + x 3 / 2 ) dx 



2lT [8x - | x 3 / 2 - x 2 + | x 5 / 2 ] J = 2. (32 



64 



16 + 



64 \ 



2^ 



§(240-320 + 192)= f (112) 



224:. 
15 



(d) V=/ c d 27T( 



shell \ / shell 
radius / I height 



dy = ^2.(2 - y) (y 2 ) dy = 2. Jj2y 2 - y 3 ) dy = 2. [§ y 3 - £] 



2tt 



16 16\ 



32. 
12 



(4-3) 



8^ 
3 



29. (a) V = J> (** ) (** ) dy = J>y(y - y 3 ) dy 

= />(y 2 -y 4 )dy = 2.[^-^ = 2.(i-i) 



4- 
15 



shell \ / shell 



(b) V = £2-( r X)^; g h t Jdy 
= / o 1 2.(l-y)(y-y 3 )dy 



x=y-y 3 




2-X'(y-y 2 -y 3 + y 4 )dy = 2-p-^-? + fl; = 2^(i-i-i + i) = |(30-20- 15+12)= |g 



30. (a) V = XM^)(^ t )dy 

= £ 2.y [1 - (y - y 3 )]dy 

= 2.j;'(y-y 2 + y 4 )dy = 2tt[£-£ + £ 



2.(1-1+1) 

117T 

15 



2^r 
30 



(15- 10 + 6) 



x=y-y 3 




(b) Use the washer method: 



V = /V[R 2 (y)-r 2 (y)]dy = X 1 .[l 2 -(y-y 3 ) 2 ]dy = 7rj; ) 1 (l-y 2 -y G + 2y 4 )dy = .[ y -^-^ + f]^ 



. 1 



^(105-35- 15 + 42) 



97- 

105 



(c) Use the washer method: 

V = £. [R 2 (y) - r 2 (y)] dy = £ . [[1 - (y - y 3 )] 2 - 0] dy = tt£ [l - 2 (y - y 3 ) + (y - y 3 ) 2 ] dy 

= SI ( 

= jfo (70 + 30 + 105 - 2 • 42) 

(d) V = J> (** ) (**) dy = £ 2.(1 - y) [1 - (y - y 3 )] dy = 2. £(l - y) (1 - y + y 3 ) dy 



(1 + y 2 + y G - 2y + 2y 3 - 2y 4 ) dy = .[y+f + ^-y 2 + f 

1217T 

210 



.(l + I + I-l+I-§) 



2 -J> 



- y + y 3 - y + y 2 - y 4 ! 



dy = 2ir£ (1 - 2y + y 2 + y 3 - y 4 ) dy = 2. v - ^ 



r 1 r _ r 
3 ' 4 5 



2.(1-1 + 1 + 1-1) 



|(20+15-12)- : ^ 



30 



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Section 6.2 Volume by Cylindrical Shells 377 



31. (a) V = J> (** ) (**)dy = /; 2*y (^ - y») dy 

[4}/2 v 5/2 



27r/ o (2v^y 3 /2 _ y a) dy = 27r [i*2 y ^ _ t 



2tt 



M 



2..4(f-l) 



_ ?1 

8tt 
5 



27T 



(8 - 5) 



f 4-2 3 _ 44 \ 
V 5 4 ) 



24- 
5 




W V = J> (*- ) ( -J) dx = /; 2.x (,/i - f ) dx = 2nj; (x 3 / 2 - f ) dx = 2n [§ x 



4 

5/2 



3 _ tt-2 4 -3 _ 48;r 
5 5 



32. (a) V=J>(,X) (**)<* 

= J" 2ttx [(2x - x 2 ) - x] dx 

= 2tt J x (x - x 2 ) dx = 2?r J (x 2 - x 3 ) dx 



2tt 



2tt 



:*-« 



(b) V=/ a b 27T( 



shell \ / shell 
radius / I height 



dx = J 2tt(1 - x) [(2x - x 2 ) - x] dx = 2tt J (1 - x) (x - x 2 



dx 



2. J o (x - 2x 2 + x 3 ) dx = 2tt I 



2 3 * + 4 



2tt ' 



2^ 



2 3 ~ 4,/ 



§ (6 - 8 + 3) 



33. (a) V = f\ [R 2 (x) - r 2 (x)] dx = tt f 16 (x~ 1/2 - l) dx 
= vr [2xV2 _ x ] J /i6 = vr [(2 - 1) - (2 - I - ^)] 



7T (1 



6) 



9^ 
16 



W V = X 2. (** ) ( -J) dy = J>y (£ - *) dy 



1 v" 2 - Z- 

2 ' 32 



if (8 + i) 



0.5 




y = i 



0^ 0:4 o.<s o:s — i - * 



9tt 
16 



34. (a) V = f\ [R 2 (y) - r 2 (y)] dy = f\ ( ^ - i) dy 

= -[-|y- 3 -^i = -[(-^-l)-(-|-^)] 

= ^(-2-6+16 + 3)=^ 

w v = £2- (*-) (-J) dx = j>x (^ - 1) dx 

= 2 -/> i/2 - x ) dx =M§* 3/2 -C 



y 

2- 

i- 




y-1/Vx 



.25 



1/4 



= 2 ^[(5-5)-(M-^)]=^(5- 1 -5+ii;) = 4|( 4 - 1 6-48-8 + 3) 

35. (a) Disk: V = Vi - V 2 

Vi = f V[Ri(x)] 2 dx and V 2 = f 2 7r[R 2 (x)] 2 with Ri(x) = A /^~ and R 2 (x) = Jx, 

J ai J ;i2 V 

ai = —2, bi = 1; a 2 = 0, b 2 = 1 => two integrals are required 

(b) Washer: V = V, - V 2 

V! = J% ([Ri(x)] 2 - [ ri (x)] 2 ) dx with R x (x) = 



11- 

48 



^±2 and ri ( x ) = 0; ai = -2 and bi = 0; 



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32 




378 Chapter 6 Applications of Definite Integrals 



V 2 = / ' 7T ([R 2 (x)] 2 - [r 2 (x)] 2 ) dx with R 2 (x) = y^ and r 2 (x) = y/l; a 2 = and b 2 = 1 

=> two integrals are required 

(c) Shell: V = £ 2n (** ) ( -J ) dy = £ 27 ry ( **) dy where shell height = f - (3y 2 - 2) = 2 - 2y 2 ; 

c = and d = 1. Only one integral is required. It is, therefore preferable to use the shell method. 
However, whichever method you use, you will get V = n. 



36. (a) Disk: V = Vi - V 2 - V 3 

Vj = J 7r[Ri(y)] 2 dy, i = 1, 2, 3 with Rj(y) = 1 and ci = -1, di = 1; R 2 (y) = ^/y and c 2 = and d 2 = 1; 

R3(y) = (~ y) 1 ^ 4 an d C3 = — 1, d3 = =>■ three integrals are required 

(b) Washer: V = Vi + V 2 

V s = J\([R,(y)] 2 - [r(y)] 2 ) dy, i = 1, 2 with Ri(y) = 1, n(y) = y/y, Ci = and di = 1; 
R 2 (y) = 1> r 2 (y) = (— y) 1 ' 4 , c 2 = — 1 and d 2 = => two integrals are required 

(c) Shell: V=/>(^j(^ t )dx =/>x( h s e 1 ;: i : t )dx, where shell height = x 2 -(-x^) = x 2 + x 4 , 
a = and b = 1 => only one integral is required. It is, therefore preferable to use the shell method. 



However, whichever method you use, you will get V 



6 • 



6.3 LENGTHS OF PLANE CURVES 



1. | = _ land £ = 3 



Length = J^a/iOA = v/lO [t] 



1-2/3 



fVio 



10 



5/10 



2. 



I = -sintand| = l+cost ^ J(f) 2 + (%)* = ^C" sin t) 2 + (1 + cos t) 2 = ^2 + 2 cos t 

=> Length = / o V2 + 2cost dt = ^2 fiy/fc***) (1 + ^0 dt = v^ /J^^ dt 

= \/2 I — ^-SiSJ — dt (since sin t > on [0, 7r]); [u = 1 - cos t => du = sin t dt; t = => u = 0, 

v Jo s/\ -cost v — ' y 

t = 7T =>■ u = 2] -> a/2 /* u" 1 /' 2 du = a/2 [2U 1 / 2 ] * = 4 



3. f =3t 2 and£=3t 



dx\ 2 
, dtJ 



(I) = ^/(3t 2 ) 2 + (3t) 2 = ^9t 4 + 9t 2 = 3t Vt 2 + 1 (since t > on [o, v^] ) 



Length = J q St^/t 2 + 1 dt; [u = t 2 + 1 => § du = 3t dt; t = =>• u = 1, t = \/3 =>• u 

J i 4 |u 1 / 2 du= [u 3 / 2 ] 4 = (8 - 1) = 7 



dx 



dy 



4. f=tandf = (2t+l) 1 /2 => J(^) 2 4-(^) = Vt 2 + (2t 4- 1) = ^0+1)^ = |t 4- 1| = t + 1 since < t < 4 



Length = J^(t + 1) dt = [f + t] = (8 4- 4) = 12 



5. |=(2t + 3) 1 / 2 and^ = l+t =► ^ (f) 2 + {%)" = y/Qx + 3) + (1 + t) 2 = ^t 2 4- 4t 4- 4 = |t + 2| = t + 2 
since < t < 3 => Length = J fl (t 4- 2) dt = [| + 2tl 



2J_ 
2 



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Section 6.3 Lengths of Plane Curves 379 



6. f t = 8t cost and ^ = 8t sin t => J(f ) 2 + (^) = v/(8t cos t) 2 + (8t sin t) 2 = ^64t 2 cos 2 1 + 64t 2 sin 2 1 



|8t| = 8t since < t < § =4> Length = f ~8t dt = [4t 2 ] 



" /2 - tt 2 



dy _ I . 3 , v 2 , n1/2 



dx 3 2 



(x 2 + 2) V -2x= v/(x 2 +2)-: 



L = J v/l + (x 2 + 2)x 2 dx = JVl + 2x 2 +x 4 dx 



J o y(l+x 2 ) 2 dx=X(l+x 2 )dx=[x+f] o 



3 + f = 12 




-I — X 



dy _ 3 
dx 2 



L =X 4 V /r +^ dX; [ U=1 + 3 X 



du = | dx =>• | du = dx; x = => u=l;x = 4 



=> u = 10] -> L 



/■io 
, uV 2 



du 



[| U 3/ 2 ]; 



10 



27 



(lo^/To- l) 




Q ^ 2 

7 - dy y 4y 



,4 lit 



y 2 ^ (dy) 



2 ' 16y 4 



/,Vy 4 + 5 + i^ d y 



J,V(y 2 + ^)" d y = r(y 2 + ^) d y 




'27_I1_fl_lUQ_I_l i i-q i (-1-4 + 3) _ n , (-2) _ 53 



3 12-/ V3 4/ 



12 3 ' 4 



io- | = |y 1/2 -b- 1/2 ^ (l) 2 = K y - 2 + rt 



=*L=j; ^l + i(y-2+i)dy 

= J,VK y+2+ O dy = -T* vW*)' d y 

= |/ 9 (y 1/2 + y" 1/2 ) dy = 5 [| y 3/2 + 2 y 1/2 ] J 



,3/2 



+ 



y 1/2 ]; = (f + 3)-G + i) 



ii 



1 _ 32 
3 — 3 



9- 


y#2 


<■ 






5 



11. 



Jy 



- V 3_J_ =v fdx\ 2 6_1 , J_ 

— y 4y3 ^ \dy) - y 2 + 16y 6 

>L =r v /i+y6 -5 + T^ dy 

J.y y° + 5 + T^ dy = /; y(y 3 + ^) 2 dy 



J/( y3 + ^) dy 

fie L_"\ _ (i-l) 



y_ _ y_ 

4 8 



1 1 , 1 _ 128-1-8+4 _ 123 



32 4 



32 



32 




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380 



12. p 

dy 



Chapter 6 Applications of Definite Integrals 

= £-£ => (|) 2 = Hy 4 -2 + y - 4 ) 

> L = Xyi + |(y 4 -2 + y^)dy 
J 2 yi(y 4 + 2 + y-4)dy 

^/;V(y 2 + y^ 2 ) 2d y = I IV + y~ 2 ) d y 




1 r 

2 3 



I [(f - 1) - (! - 1)] = Hf - ! + 1) = 1 (6 + 1) 



13 

4 



13. 



| =x l/3_I x -l/3 ^ 

^ L =/ 1 V i + x2/3 



m 



X 

1 i X-2/3 

2 ' 16 



2/3 _ 1 
2 



t-2/3 
l6~ 



dx 



dx 



J 1 V x2/3 +5+ 

/^(xVS-f l x -l/3) 2 dx = /'(xVS + 1 X-V3) dx 
r| x 4/3 + | x 2/3l8 3 r 2x 4/3. 



,2/31 



3 [( 2 . 2 4 + 2 2 ) - (2 + 1)] = | (32 + 4 - 3) = f 



20" 








15 


y = 


3x 4 ' 3 
4 


3x2,3 « 
8 ^/ 




10 










5 


1 






H H»-X 



14- t 



x 2 + 2x + 1 - 



(4x+4) 2 



X 2 + 2x + 1 - i ! 



4 (1+x) 2 



(l+-) 2 -?(IW => (I) =d+*) 4 -2- + T6(lW 






l+x) 



2 i (1+x)- 




dx; [u = 1 + x => du = dx; x = => u = 1, x = 2 =^ u = 3] 






108-1-4+3 _ 106 _ 53 
A) ~ " 12 12 6 



15. 



dx 
dy 



V^ec 4 y-1 => (|) 2 = sec 4 y-l 

> L = J^ /4 v/l + (sec 4 y-l) dy = J^sec 2 y dy 
[tany]^ /4 =!-(-!) = 2 



..x= I Vsec 4 t -1 dt 




-91 /4- 



->— x 



16. 



dy 



Vi> 



m 



3x 4 -l 



=> L = f_' v/l + (3x 4 - 1) dx = /_,' a/3 x 2 dx 



'3 If 



v/3 



l-(-2) 3 



i/3(_l + 8) = ^ 



-2 



y=JV3t 4 -1dt 

-2 



-1 



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Section 6.3 Lengths of Plane Curves 38 1 



17. (a) g=2x^ (I)' =4x^ 



- L =fY i+ (i0 2dx 

/^ v 7 ! +4x 2 dx 



(c) L«6.13 



(b) 




i — -o'.s b 0:5 1 ■ ■ i:j ■ ■ ■ t- ■ 



18. (a) g = sec 2 x => (g) 2 = sec 4 x 
=>■ L = J ^ a/ 1 + sec 4 x dx 



(c) L«2.06 



(b) 



-] -0.8 -0.6 -0.4 -0.2 




19. (a) I = cosy => (^J = cos 2 y 

=> L = I a/1 + cos 2 y dy 
(c) L « 3.82 



(b) 




20. (a) 



dx 



dy v/i^^y 5 



(*)' 



y 

l- y 2 



L " J-I/2V ! + I 1 ^) 

X!>-yT 1/2 dy 



^JL/S^y 



(c) L« 1.05 



(b) 



x-vr^F 




NOT TO SCALE 



21. (a) 2y + 2 = 2| => (|) 2 = (y+l) 2 

=> L = £ iV /l + (y+l) 2 dy 
(c) L«9.29 



(b) 



y 2 + 2y = 2x + 1 



rm-rrr ' 



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382 Chapter 6 Applications of Definite Integrals 

22. (a) -Z- = cos x - cos x + x sin x =>- ( ■£■ ] = x 2 sin 2 x 

=> L = J \/l + x2 sin 2 x dx 
(c) L«4.70 



(b) 




i!> 



2.1 (a, I=tanx^ (g)' 



o 

(c) L«0.55 



L = J \/l +tan 2 x dx = J 
-S 25 - = sec x dx 

Jo cos x Jo 



tt/6 



sin^ x + cos^ x 



(b) 



dx 




y OM D2 53 53 5T~ * 



24. (a) ! = v /sec 2 y-l => (|) 2 = sec 2 y-l 
=► L = J^ /3 v/l + (sec 2 y-l) dy 

Xir/4 r>7r/4 

|secy|dy= J secydy 



(c) L«2.20 



(b) 



x= [ vsec 2 1-1 dt 
Jo 



25. 



V^x = / o yi+(^) 2 dt,x>0^ v / 2"=yi + (^) 1 



real number. 



dy 
dx 



± 1 => y = f(x) = ± x + C where C is any 



26. (a) From the accompanying figure and definition of the 

differential (change along the tangent line) we see that 
dy = f'(x k _,) Ax t 4 length of kth tangent fin is 
V( A x k ) 2 + (dy) 2 = y/( A x k ) 2 + [f'(x k _,) A x k ] 2 . 



(■h-vf^k.i 



y=/Cx), 




Tangent fin 
with slope 

A*.,) 



(b) Length of curve = lim J] (length of kth tangent fin) = lim J] a/( A x k ) 2 + [f'(x k _,) A x k ] 2 

k=l k=l 

= n lim oo £ Vl + [f'(x k -i)] 2 Ax k = / a Vl + [f'W] 2 dx 



27. (a) ( jj| ] correspondes to i here, so take ^ as -^-. Then y = y^ + C and since (1, 1) lies on the curve, C = 0. 
So y = y/x from (1, 1) to (4, 2). 
(b) Only one. We know the derivative of the function and the value of the function at one value of x. 

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Section 6.3 Lengths of Plane Curves 383 



28. (a) (p-) correspondes to X here, so take ^ as \. Then x = — - + C and, since (0, 1) lies on the curve, C = 1 

Soy =T ^. 
(b) Only one. We know the derivative of the function and the value of the function at one value of x. 



29. (a) f = -2sin2tand^f = 2 cos 2t => J (f t ) 2 + (f) = ^{-2 sin 2t) 2 + (2 cos 2t) 2 = 2 
=> Length = JJ 2 2 dt = [2t] l' 2 = tt 



(b) f = tt cos 7rt and f 



-7T Sin 7Tt 



V S ) 2 + (^) 2 = V 7 ^ COS 7rt)2 + (-7T Sin 7rt)2 = TT 

=>• Length = J 7r dt = [7rt] _ x , 2 = tt 



30. x = a(0 - sin 0) 

dy 



^ °| = a sin =^ ( ^ 



(8) 



a(l — cos 0) 

2 



'dx\ 2 



a 2 (1 — 2 cos + cos 2 0) and y = a(l — cos 0) 

•>2tt 



a 2 sin 2 => Length = £ * J (|) 2 + (|) d0 = XV 2 ^ 1 - cos 0) d0 
V^/r^T 11 ! 2 ^^ = 2a £>in f | d0 = 2a /"sin f d0 = -4a [cos f] J" = 8a 



31-36. Example CAS commands: 
Maple : 

with( plots ); 

with( Student[Calculusl] ); 

with( student ); 

f := x -> sqrt(l-x A 2);a := -1; 

b:= 1; 

N := [2, 4, 8 ]; 

for n in N do 

xx := [seq( a+i*(b-a)/n, i=0..n )]; 

pts := [seq([x,f(x)],x=xx)]; 

L := simplify(add( distance(pts[i+l],pts[i]), i=l..n )); 

T := sprintf("#31(a) (Section 6.3)\nn=%3d L=%8.5f\n", n, L ); 

P[n] := plot( [f(x),pts], x=a..b, title=T ): 
end do: 

display( [seq(P[n],n=N)], insequence=true, scaling=constrained ); 
L := ArcLength( f(x), x=a..b, output=integral ): 
L = evalf( L ); 

37-40. Example CAS commands: 
Maple : 

with( plots ); 

with( student ); 

x:=t->t A 3/3; 

y := t -> t A 2/2; 

a:=0; 

b:=l; 

N := [2, 4, 8 ]; 

for n in N do 

tt := [seq( a+i*(b-a)/n, i=0..n )]; 

pts := [seq([x(t),y(t)],t=tt)]; 



#(b) 
#(a) 

#(c) 



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384 Chapter 6 Applications of Definite Integrals 

L := simplify(add( student[distance](pts[i+l],pts[i]), i=l..n )); # (b) 

T := sprintf("#37(a) (Section 6.3)\nn=%3d L=%8.5f\n", n, L ); 

P[n] := plot( [[x(t),y(t),t=a..b],pts], title=T ): # (a) 

end do: 

display( [seq(P[n],n=N)], insequence=true ); 

ds := t ->sqrt( simplify(D(x)(t) A 2 + D(y)(t) A 2) ): # (c) 

L := Int( ds(t), t=a..b ): 
L = evalf(L); 

31-40. Example CAS commands: 

Mathematica : (assigned function and values for a, b, and n may vary) 
Clear[x, f] 

{a, b} = {-l, l};f[x_] = Sqrt[l-x 2 ] 
pl=Plot[f[x], {x, a, b}] 
n = 8; 

pts = Table[{xn, f[xn]}, {xn, a, b, (b - a)/n}]// N 
Show[{pl,Graphics[{Line[pts]}]}] 

Sum[ Sqrt[ (pts[[i + 1, 1]] - pts[[i, l]]) 2 + (pts[[i + 1, 2]] - pts[[i, 2]]) 2 ], {i, 1, n}] 
NIntegrate[ Sqrt[ 1 + f [x] 2 ],{x, a, b}] 

6.4 MOMENTS AND CENTERS OF MASS 

1 . Because the children are balanced, the moment of the system about the origin must be equal to zero: 
5 . 80 = x • 100 =^ x = 4 ft, the distance of the 100-lb child from the fulcrum. 



2. Suppose the log has length 2a. Align the log along the x-axis so the 100-lb end is placed at x = — a and the 

x = f . That is, x is located 



100(-a) + 200(a) 



300 



200-lb end at x = a. Then the center of mass x satisfies x 

at a distance a — I = y = i (2a) which is I of the length of the log from the 200-lb (heavier) end (see figure) 

2 



or j of the way from the lighter end toward the heavier end. 



i(2a) 



100 lbs. 



a/3 



200 lbs 



The center of mass of each rod is in its center (see Example 1). The rod system is equivalent to two point 
masses located at the centers of the rods at coordinates ( | , 0) and (0, |) . Therefore x = — 



_ x 1 m 1 +x 2 m2 

m l+ m 2 

mass location. 



5- m +0 
m+m 



j andy 



yim 2 +y 2 m 2 
m!+m 2 



0+|-m 
m+m 



f j, ¥) is the center of 



4. Let the rods have lengths x = L and y = 2L. The center of mass of each rod is in its center (see Example 1). 
The rod system is equivalent to two point masses located at the centers of the rods at coordinates (fe, 0) and 



(0, L). Therefore x 



|-m+0-2m 
m+2m 



!andy=2^n 



2L 
3 



Ji > t) * s tne center of mass location. 



5. M = / o 2 x-4dx= [. 



4f 



4-3 



8; M = f\ dx = [4x] 2 =4-2 = 8 



Mo _ i 

M L 



6. Mo = J] x • 4 dx = U f 1 = \ (9 - 1) = 16; M = f } 4 dx = [4x]f =12-4 = 8 



Mo 
M 



16 

8 



Copyright (c) 2006 Pearson Education 




Dt 



7. M, 



o 



I^ 1 



+ f)dx 



/„'( 



3 + 



Mo 
M 



:/ x +f)d X =[f+fi; 



15_5 
9 — 3 



Section 6.4 Moments and Centers of Mass 385 

f;M = X 3 (i + i)dx=[ x + f 



'9 , 27 
,2 ' 9 



M o=r x (2-t) dx =r( 2x -T)d X =[x 2 



16 



64 \ 



16 -f 



16- ?- = ^ 



Mn 

M 



32 _ 16 
3-6 9 



~1T 



f) 



2^3/ 



9. M = / i 4 x(l + -l ; )dx = / i 4 (x + X V 2 )d X =[| 

M = J^l + x- 1 / 2 ) dx = [x + 2X 1 / 2 ] J = (4 + 4) - (1 + 2) = 5 => x=^ = ^ 



15 , 14 _ 45+28 _ 73 . 
2 ' 3 — 6 — 6 ' 



73 
30 



10. M = J i/4 x - 3 (x- 3 / 2 + X-V2) dx = 3/ i/4 (x-V2 + x -3/ 2) dx = 3 [2x i/2 _ _2_] J /4 = 3 [(2 - 2) - (2 - I 



U> 



3(4 - 1) = 9; M = 3 r 4 (x-3/ 2 + x^/ 2 ) dx = 3 [£ - ^] J = 3 [(-2 - |) - (-4 - f )] = 3 (2+ f) 



6 + 14 = 20 



1/4 

Mo 
M 



20 



11. M = J o x(2 - x) dx + Jj x - x dx = f Q (2x - x 2 ) dx + f'x 2 dx = [ 
= £ = 3; M = J o (2 - x) dx + Jj x dx = 1 2x 



2x 2 x 3 
2 3 



1 







1 



2- 2 ) 



+ 



v2 2/ 



;i-d + (!-^ 



M^ 
M 



12. M = / o 'x(x+l)dx + / i 2 2xdx=/ o 1 (x 2 + x)dx + / i 2 2xdx= [f + f ] + [x 2 ]i = (5 + 5) +(4-1) 



3 + f = f ; M = / o '(x + 1) dx + \\ dx = [f + x] * + [2x] 2 = (i + l) + (4 - 2) = 2 + § 



M. 



;i)(i; 



23 
21 



13. Since the plate is symmetric about the y-axis and its density is 
constant, the distribution of mass is symmetric about the y-axis 
and the center of mass lies on the y-axis. This means that 
x = 0. It remains to find y = ^ • We model the distribution of 
mass with vertical strips. The typical strip has center of mass: 
Cx ,y ) = I x, ^^ ) , length: 4 — x 2 , width: dx, area: 
dA = (4 — x 2 ) dx, mass: dm = 6 dA = 6 (4 — x 2 ) dx. The moment of the strip about the x-axis is 
y dm = ( 2L y^ j 6 (4 — x 2 ) dx = | (16 — x 4 ) dx. The moment of the plate about the x-axis is M x = J Ij dm 




= £j(16-x 4 )dx=f[l6x-^] 2 _ 2 =f[(l6-2 
plate is M = / 6 (4 - x 2 ) dx = 6 Ux - f 1 " = 26 (8 



0-(-l«-2 + f)] 



^P . Therefore y 



<5-2 



Mx 
M 



32 



f) 



!!&•■ 



The mass of the 



«! 



The plate's center of 



mass is the point (x, y) 



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386 Chapter 6 Applications of Definite Integrals 



14. Applying the symmetry argument analogous to the one in 



Exercise 13, we find x = 0. To find y 



M 



we use the 



y=25-x2 



vertical strips technique. The typical strip has center of 

mass: (x ."y ) = ( x, 25 ~ x j , length: 25 — x 2 , width: dx, 

area: dA = (25 - x 2 )dx, mass: dm = 8 dA = S (25 - x 2 ) dx. 
The moment of the strip about the x-axis is 

2S 2 ~ x2 J S (25 - x 2 ) dx = | (25 - x 2 ) 2 dx. The moment of the plate about the x-axis is M x = jy dm 
f_ 5 f (25 - x 2 ) 2 dx = | f s (625 - 50x 2 + x 4 ) dx = g [625x - 




y dm 



2 



50 v 3 J- * 
3 X -t- 5 



50 



2 - f (625 • 5 - f ■ 5 3 



i) 



6 - 625 (5 - f + 1) = 6 ■ 625 - (§) . The mass of the plate is M = J dm = / \. <5 (25 - x 2 ) dx = 6 [25x - 



26 15 



^ 8 ■ 5 . Therefore y 



Mi 

M 



,5.54. (M 

3 h\ = 10. The plate's center of mass is the point (x, y) 



(0, 10). 



15. Intersection points: x — x 2 



2x 







x(2 — x) = => x = 0orx = 2. The typical vertical 

(x-x 2 ) + (-x) N 



y=x-x 2 




y=-x^^^^ 


2 



strip has center of mass: ( x , y ) = I x, 

= (x, - y J , length: (x - x 2 ) - (-x) = 2x - x 2 , width: dx, 

area: dA = (2x — x 2 ) dx, mass: dm = 6 dA = 8 (2x — x 2 ) dx. 
The moment of the strip about the x-axis is 

y dm = ( — y J 8 (2x — x 2 ) dx; about the y-axis it isx" dm = x • 8 (2x — x 2 ) dx. Thus, M x = J y dm 

= - X 2 (I x ' 2 ) ( 2x - x2 ) dx = - iX 2 ( 2x3 - x4 > dx = - f [^ - f ] = - 1 ( 23 - f ) = - 1 - 23 (1 

= - f ; M y = / x dm = f\ ■ 8 (2x - x 2 ) dx = ^J^x 2 - x 3 ) = 8 [| x 3 - £| ' = 8 (l 2 * ~ ' 



-1) 




S-2 4 
12 


3 



M 



/ dm = £ 8 (2x - x 2 ) dx = sf*(2x - x 2 ) dx = 6 \\ 2 



6(4 



4h 



3 . Therefore, x = ^ 



\ 3 ) \4Sj 



1 and y 



M, 

M 



¥)(4) 



(x, y) = (l, — |) is the center of mass. 



16. Intersection points: x 2 — 3 = — 2x 2 =>■ 3x 2 — 3 = 

=^ 3(x - l)(x + 1) = => x = — 1 or x = 1. Applying the 
symmetry argument analogous to the one in Exercise 13, we 
find x = 0. The typical vertical strip has center of mass: 

(x ; y)=(x, - 2 " 2+ f- 3 ) ) = (x,^, 

length: -2x 2 - (x 2 - 3) = 3 (1 - x 2 ), width: dx, 

area: dA = 3 (1 — x 2 ) dx, mass: dm = 8 dA = 38 (1 — x 2 ) dx. 

The moment of the strip about the x-axis is 




y=x*-3 



y dm = \ 6 (-x 2 - 3) (1 - x 2 ) dx = | 8 (x 4 + 3x 2 - x 2 - 3) dx = | 8 (x 4 + 2x 2 - 3) dx; M x = J y dm 



Sfjx* + 2x 2 - 3) dx = \6 [f + ^ - 3x 



M = / dm = 3(5^(1 - x 2 ) dx = 38 |x - 
=>- (x, y ) = (0, — |) is the center of mass. 



l-*-2(| 



38-2(1-1) 



-3 



38 



'3+ 10-45 i 



3 -V ~~ -'" V 15 

46. Therefore, y = 



326 . 
5 ' 



6-32 
5-<5-4 



Copffigl (c| 1 Pearson Educate k, publishing as Pearson Addison-Wesle 



Section 6.4 Moments and Centers of Mass 387 



17. The typical horizontal strip has center of mass: 

Cx ^ ) = ( ^-ir-,y ) , length: y — y 3 , width: dy, 

area: dA = (y — y 3 ) dy, mass: dm = 8 dA = 8 (y — y 3 ) dy. 
The moment of the strip about the y-axis is 



x dm 



< 5 ( I ^)(y-y 3 ) d y = f(y-y 3 ) 2d y 

(y 2 — 2y 4 + y 6 ) dy; the moment about the x-axis is 




y dm = <5y (y — y 3 ) dy = 6 (y 2 — y 4 ) dy. Thus, M x = J y dm = 8 I (y 2 — y 4 ) dy = <5 

M y = /xdm=fX 1 (y 2 -2y 4 + y«)dy=f[f- 2 f + ^ 

y 



8 f 35 -42 +15 ' 
2 V 3-5-7 / 



hi-i: 



28 
15 



48 
105 



; M = / dm 



*xv 



dy = 8 \- 



ii 



Hl-l) 



Therefore, x 



_8_ 
15 



(x,y) 



16 8 



is the center of mass. 



My 

M" 



' 48 ' 

,105; 



16 
105 



and y 



M 



1) 



18. Intersection points: y = y 2 — y =>■ y 2 — 2y = 
=> y(y — 2) = =>• y = 0ory = 2. The typical 
horizontal strip has center of mass: 

(x,y)=(^j) = p,y), 

length: y — (y 2 — y) = 2y — y 2 , width: dy, 
area: dA = (2y — y 2 ) dy, mass: dm = 8 dA = 6 (2y — y 2 ) dy. 
The moment about the y-axis is3c dm = § • y 2 (2y — y 2 ) dy 
= | (2y 3 — y 4 ) dy; the moment about the x-axis is y 1 dm = <5y (2y — y 2 ) dy = 8 (2y 2 — y 3 ) dy. Thus, 




M x = Jy dm = I 8(2y 2 - y 3 ) dy = 8 [f - 1 

= Tl ( 2 y 3 - y 4 ) d y = I [^ - f ] o = i ( 8 - 

= 8 [y 2 - f 1 2 = 5 (4 - |) = f . Therefore, x 



*( 



¥) 



*5 (4 - 3) = f ; M y = / x dm 



16-' 



f( 40_32; 



46 . 



; M = J dm = f 8 (2y - y 2 ) dy 



(x,y) 



M, 



f4«' 

v 5 / 



y4SJ 



and y 



, l) is the center of mass. 



Ma 

M 



:f ) (*) 



19. Applying the symmetry argument analogous to the one used 
in Exercise 13, we find x = 0. The typical vertical strip has 



center of mass: (x ,y ) 



, length: cos x, width: dx, 



y=cos x 



area: dA = cos x dx, mass: dm = 8 dA = 8 cos x dx. The 
moment of the strip about the x-axis is y dm = 8 ■ ^p • cos x dx 



| cos 2 x dx 



8 ( 1 + cos 2x *\ 



dx 



f (1 + cos 2x) dx; thus 
M , = J y dm = J*'* £ (1 + cos 2x) dx = 



8 [sin x 



if/2 

I -tt/2 



2 V 2 

7T/2 

2<5. Therefore, y = ^ 




-*/2 



x + ^1 ' 1 = I [(f + 0) - (- §)] = f ; M = / dm = 8 /* ms x dx 



i-28 



8 



2 J -jr/2 



20. Applying the symmetry argument analogous to the one used 
in Exercise 13, we find x = 0. The typical vertical strip has 

center of mass: ( x ,y ) = I x, ^L* J , length: sec 2 x, width: dx, 

area: dA = sec 2 x dx, mass: dm = 8 dA = 8 sec 2 x dx. The 

moment about the x-axis is y* dm = ( ^^ J (8 sec 2 x) dx 

, y dm = I I . 



(0, 1) is the center of mass. 

y 



/*7r/4 /W 4 

sec 4 x dx. M. = | y dm = £ | sec 4 x 

5 J-jt/4 ^ 2 J_a-M 



dx 




-jt/4 



(*> y) km 



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388 Chapter 6 Applications of Definite Integrals 



/^(tan 2 x + 1) (sec 2 x) dx = § f^\ (tan x) 2 (sec 2 x) dx + f J^sec 2 x dx = f [^ 



-w/4 



I[tanx]t /4 



« ri 

2 L3 



Therefore, y 



I)] 



M, 



}[!-(-!)] 



;M= fdm=<5 P sec 2 x dx = <S[tan xf /4 ,, = <5[1 - (-1)] 

J J —71/4 7T/4 



'^ (J_ x 

\ 3 J V2*/ 



(x,y) 



(0, 1) is the center of mass. 



26. 



21. Since the plate is symmetric about the line x = 1 and its 
density is constant, the distribution of mass is symmetric 
about this line and the center of mass lies on it. This means 
that x = 1. The typical vertical strip has center of mass: 

t-x 2 )+(2x 2 -4x) 



(x ,y ) 



/ (2x-x 2 ) + {2x 2 -4x) \ 

V x ' 2 ) 



length: (2x - x 2 ) - (2x 2 - 4x) 



3x 2 + 6x = 3 (2x - x 2 ) , 



width: dx, area: dA = 3 (2x — x 2 ) dx, mass: dm = 6 dA 
= 3<5 (2x — x 2 ) dx. The moment about the x-axis is 

y dm = | 6 (x 2 - 2x) (2x - x 2 ) dx = - § 6 (x 2 - 2x) 2 dx 




y = 2x i - Ax 



- \6 (x 4 - 4x 3 + 4x 2 ) dx. Thus, M, = / y dm = -f o § <5 (x 4 - 4x 3 + 4x 2 ) dx = - § 6 [f - : 
-^(f-2 4 + 4 .23)=-^.2 4 (|-l + |)=-^.2 4 (^±i0)=-f;M=/dm 

£36 (2x - x 2 ) dx = 36 \r 
■ (x,y) 



. -: . . v " = 36 (4 - § ) = 46. Therefore, y = f = (- f ) (£) 



'l, — I) is the center of mass. 



4 3 

3 * 



22. (a) Since the plate is symmetric about the line x = y and 
its density is constant, the distribution of mass is 
symmetric about this line. This means that x = y. The 
typical vertical strip has center of mass: 

(x ,y ) = (x, ^p") , length: ^9 - x 2 , width: dx, 

area: dA = \J 9 — x 2 dx, 

mass: dm = 6 dA = 6y 9 — x 2 dx. 

The moment about the x-axis is 



2+y2=9 




ydm = « (^p^) ^9-x 2 dx = f (9 - x 2 ) dx. Thus, M x = J y dm = £ § (9 - x 2 ) dx = | [9x - y] 
= § (27 -9) = 96;M = Jdm= J 6 dA = 6 J dA = <5(Area of a quarter of a circle of radius 3) = 6 ( 9 -f) 



9ttS 
4 



Therefore, y = % = (96) (^) = % => (x,y) 



is the center of mass. 



(b) Applying the symmetry argument analogous to the one 
used in Exercise 13, we find that x = 0. The typical 
vertical strip has the same parameters as in part (a). 

Thus, M x = fy dm = J" § (9 - x 2 ) dx 

= 2 £ f (9 - x 2 ) dx = 2(9,5) = 18,5; 
M = /dm=/5dA = ( 5 /dA 

= 6(Area of a semi-circle of radius 3) = 6 {^f 
as in part (a) =^> (x, y) = (0, -) is the center of mass. 



2 



5sx2+y2=9 




- , the same y 

7T ' J 



Copyright (c) 2006 Pearson Education 




Section 6.4 Moments and Centers of Mass 389 



23. Since the plate is symmetric about the line x = y and its 
density is constant, the distribution of mass is symmetric 
about this line. This means that x = y. The typical vertical 
strip has 
center of mass: (x ,y ) = (x, 3 +V 2 9 ~ x j > 

length: 3 — v 9 — x 2 , width: dx, 
area: dA = (3 - a/9 - x 2 ) dx, 

mass: dm = 6 dA = 6 (3 — v 9 — x 2 j dx. 

The moment about the x-axis is 

^ (3 + N /9^?)(3-v , 9^^) 
y dm = ^ '- dx 




= I (4 - 7T) = 

center of mass. 



[9 - (9 - x 2 )] dx 
with side length 
M = SA = f (4 - tt). Therefore, y 



<\x- 



dx. Thus, M, 



f 

Jo 



§[x 3 ] 3 = f . The area 



equals the area of a square with side length 3 minus one quarter the area of a disk with radius 3 => A = 3 2 — ^ 



'9S 

2 ) I 9(5(4 - w) 



2 

4-7T 



(x,y) 



V4— 7T ' 4 — 7T. 



is the 



24. Applying the symmetry argument analogous to the one used 
in Exercise 13, we find that y = 0. The typical vertical strip 



has center of mass: (x ,y ) 



(x,0), 



4 , width: dx, area: dA = A dx, 



length: £ _ (_ £ 

mass: dm = 8 dA = ^ dx. The moment about the y-axis is 

x dm = x • |f dx = | dx. Thus, M y = / x dm = f' 

J dm = f 



2^ 



dx 



26 



i]'i 



2/> 

2<5(a- 1) 



+ 1) 

a 2 
a- 1 - 1 



2t5(a-l) 
a 

2a 
a+1 




; M 



1 |dx = 6[-^]; 



6(-h 



(x,y) 



2a 



^-.0) . Also, lim x = 2. 

+ 1 ' / a — > 00 



<5(a 2 -l) 



Therefore, 



25. M x = /ydm=/ i if ■ 5 - ( JO dx 

= r(^)U 2 )(l)dx=r^dx = 2/;x- 2 dx 
= 2 [-x" 1 ]? = 2 [(-i) _(_!)]= 2 (I) = 1; 
M J = /xdm=J i x-«5- (J.) dx 
= J;x(x 2 )(!)dx = 2j; 2 xdx = 2[f]' 
= 2(2- ±) =4-1=3; M = /dm = J^<S ( J>) dx = J] x 2 (£) dx = 2 J]' dx = 2[x] 2 = 2(2- 1) = 2. So 

M — 2 — ' *- X > y^ — V2' 2) 







V 


=2/x2 


(*.?) 




• 


«NH| 


3 


L 




2 



x = -rf = I and y = ^ — I =>. (x, y) = (|, ^) is the center of mass 



26. We use the vertical strip approach: 

M x = Jy dm = f^-^ 1 (x - x 2 ) - £ dx 
= \f {* 2 -x 4 ) - 12xdx 
= 6£(x 3 -x 5 )dx = 6[^- 



6(1-J) 




M y = Jx dm= J\(x-x 2 )-<Sdx = J o '(x 2 -x 3 )- 12xdx= 12^'(x 3 - x 4 ) dx = 12 [^ - fj = 12 (i - ^ 



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390 Chapter 6 Applications of Definite Integrals 



I2_3 
20 5 



M 



/dm=/ (x- X 2 )-^dx=12j;'( X 2 - X 3 )d X =12[f _^]^ = 12(i-i) = {§ = 1. So 



and y 



'3 V 
v5> 2, 



is the center of mass. 



^ 1|dX 



224- 
3 



27. (a) We use the shell method: V = £ 2tt ( r ^J (^ t ) d X = f\nx [4j - (- 

= 16^/^ d X = 167r/V/ 2 d X = 16tt [| X 3 / 2 ] J = 16jt (§ - 8 - §) = ^= (8 - 1) 

(b) Since the plate is symmetric about the x-a X is and its density S(x) = - is a function of x alone, the 

distribution of its mass is symmetric about the x-a X is. This means that y = 0. We use the vertical strip 

approach to find x: M y = jx dm = J' x • ["-^ - (- 4A 1 • 6 dx = f\ - X - I dx = 8 /V 1 / 2 d X 
= 8[2 X V^ = 8(2.2-2)=16;M = /dm=j; 4 [^-(^)]^d X = 8/ i 4 (^)(I)d X = 8j;V3/ 2 d X 

= 8 [-2 X - X / 2 ] * = 8[-l - (-2)] = 8. So x = ^ = f = 2 => (x, y) = (2, 0) is the center of mass, 
(c) 



3 










l 


. 




4 




\. 4 








(2,0) 









1 


y 4 


4 




-4 











28. (a) We use the disk method: V = / ttR 2 (x) dx = J^ it (4,) dx = 4?r J x~ 2 dx = 4?r [- 1] \ 
= 4tt [^ - (-1)] = tt[— 1 + 4] = 3tt 
(b) We model the distribution of mass with vertical strips: M x = J y dm = I ^ - (|) - 6 dx = I ■% - y^c dx 

= 2/V 3 / 2 dx = 2 |"-=|] = 2[-l - (-2)] = 2; M y = Jx dm = J\ ■ \ ■ 6 dx = 2/V/ 2 dx 

= 2[^]; = 2[f-f]=f;M=/dm=/ i 42 ^dx = 2/ i 4 f dx = 2/Vv 2d x = 2[2xV^ 



(c) 



2(4 - 2) = 4. So x 



My 

M 



m 
4 



I and y 



Ma 

M 



2 _ 1 
4 — 2 



(x,y) 



'?, |) is the center of mass. 



y=2/x ,7i. 




29. The mass of a horizontal strip is dm = 6 dA = 6L dy, where L is the width of the triangle at a distance of y above 
its base on the x-axis as shown in the figure in the text. Also, by similar triangles we have jr = — ^ 



hy 2 



L = I (h - y). Thus, M s = / y dm = f* 6y (g) (h - y) dy = f Jjhy - y 2 ) dy = f [ 

= / dm =X^(K)(h-y)dy=^r(h-y)dy=^[hy-^] 



6b /V _ h^ 
h \ 2 3 



§ (h 2 - f ) 



5bh 2 (i-i; 



Sbh 2 



6bh 

2 



Soy 



M 



\ 6 y V*bhJ 



the center of mass lies above the base of the 



Copffigl (c| 1 Pearson Educate Inc., publishing as Pearson Addison-Wesle 



Section 6.4 Moments and Centers of Mass 391 

triangle one-third of the way toward the opposite vertex. Similarly the other two sides of the triangle can be 
placed on the x-axis and the same results will occur. Therefore the centroid does lie at the intersection of the 
medians, as claimed. 



30. From the symmetry about the y-axis it follows that x = 0. 
It also follows that the line through the points (0, 0) and 
(0, 3) is a median => y = | (3 - 0) = 1 => (x, y) = (0, 1). 



(0,3) 



(-1,0) 




(+1,0) 



3 1 . From the symmetry about the line x = y it follows that 
x = y. It also follows that the line through the points (0, 0) 
and (|, |) is a median => y = x=|-(|— 0)=| 
=> (x,y)=(I,i). 




(1,0) 



32. From the symmetry about the line x = y it follows that 
x = y. It also follows that the line through the point (0, 0) 



and (|,|) is a median 



2 (a 

3 V2 



o) 



(x,y) 



v3' 3/ 



(0,a) 



(0,0) 




(a,0) 



33. The point of intersection of the median from the vertex (0, b) 
to the opposite side has coordinates (0, |) 
=> y = (b - 0) - 1 = ' 
=> (x,y)=(f,|). 



| and x 



:t-o)-i 



(0,b) 



(0,0) 




(a,0) 



34. From the symmetry about the line x = | it follows that 
x = |. It also follows that the line through the points 
(f,0) and (f,b) is a median =S> y=I(b-0)=| 



(x,y) 



'a b\ 
v2' 3) 



(0,0) 




(a,0) 



35. y = x 1 / 2 =^> dy = \ x" 1 / 2 dx 
=> ds= v/(dx) 2 + (dy) 2 

Jo v 



M, 



S dx 



6 X 2 ^+T dx 



2A 
3 



;2+i 



s/2 



3/2 



f I I X 



fl\3/2 



[(!) 



1 + ^dx; 



l\3/2 



, is 3/2 



2* 111 _ II 
3 V 8 8J 



13^ 
6 




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392 Chapter 6 Applications of Definite Integrals 



36. y 



dy = 3x 2 dx 



=±> dx = J (dx) 2 + (3x 2 dx) 2 = y/l +9x 4 dx; 

M x = SJ o x 3 Vl +9x 4 dx; 

[u = 1 + 9x 4 =>■ du = 36x 3 dx =>• ^ du = x 3 dx; 

x = 0=^u=l,x=l=^u=10] 

-M x = 5 f , 3LuV 2 du=A [ | u 3/2 ] ;o = ^ (l0 3/ 2 



1 



y = x- 




2kr sin 2 C = 0; 



37. From Example 6 we have M s = f a(a sin 0)(k sin 61) d0 = a 2 k£ sin 2 d6» = & J* (1 - cos 20) 69 
= &[0- ^M] * = ^ ; M y = JJ a(a cos 0)(k sin 0) d0 = a 2 
M — I ak sin d0 = ak[— cos 0]J = 2ak. Therefore, x 
is the center of mass. 



M, 
M 



2 k f * sin cos 69 

J() -s u 

Oandy=f = (*=)(£ 



i n 



2ak/ 



o, 



38. M x = J y dm = J" (a sin 9) - <5 - a d0 
= J| ) 7r (a 2 sin^)(l + k|cos0|)d0 
= a 2 £ (sin 0)(1 + k cos 0) d0 
+ a 2 r (sin 6»)(1 — k cos 9) 69 

*J 7r/2 

= a 2 P sin 9 d<9 + a 2 k P sin cos d0 + a 2 f* sin 69 - a 2 k f" sin 9 cos d0 

Jo Jo J 7r/2 J 7r/2 




a 2 [— cos 0] 



tt/2 



Al^ 1 '" 



+ a 2 [-cos0]J /2 - a 2 k |Mn " 



2 [0 - (-1)] + a 2 k (i - 0) + a 2 [-(-l) - 0] - a 2 k (0 - \) 



k/2 

a 2 + ^ + a 2 + ^ 



= 2a 2 + a 2 k = a 2 (2 + k); 
M y = Jx dm = J*(a cos 0) • <5 • a d0 = £{a 2 cos 0) (1 + k |cos 0|) d0 

= a 2 r (cos 0)(1 + k cos 0) d0 + a 2 f (cos 0)(1 - k cos 0) d0 

JO <J 7r/2 

= a 2 /J^cos d0 + a 2 k / ( 1+c ° s2fl ) d0 + a 2 f cos d0 - a 2 kjj 2 



1 + cos20> 



d0 



a 2 [sin0] 

o2 



t/2 , a 2 k 



i2»1 7r /2 



fa , sin 29 ] "V 2 i .^Tcir, fll^ a 2 k \n , sin2ff l ff 

« 2 (1 - 0) + & [(I - 0) - (0 + 0)] + a 2 (0 - 1) - & [(tt + 0) - (f + 0)] = a 2 
M = J** <5 • a d0 = a J"(l + k |cos 0|) d0 = a /* (1+ k cos 0) d0 + &f* (1 - k cos 0) d0 



. 2 k7T „2 a 2 k7T f\. 

— - a - — - u, 



tt/2 



a[0 + k sin 0]^ + a[0 - k sin 9Y n/2 = a [(f + k) - 0] + a [(tt + 0) - (f - k)] 



f + ak + a (f + k) = a?r + 2ak = a(7r + 2k). So x 
(0, 2 T a + 2 k k a ) is the center of mass. 



and y 



M, _ a 2 (2 + k) _ a(2 + k) 
M a(7r + 2k) it + 2k 



39. Consider the curve as an infinite number of line segments joined together. From the derivation of arc 
length we have that the length of a particular segment is ds = i/(dx) 2 + (dy) 2 . This implies that 

M x = J 6y ds, M y = J 6x ds and M = J 6 ds. If 6 is constant, then x = -^ = ~y s = *-^— and 

77 _ m, _ /yds _ /yds 
' M J ds length ' 



40. Applying the symmetry argument analogous to the one used in Exercise 13, we find that x = 0. The typical 



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Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 393 



vertical strip has center of mass: (x ,y ) = I x, —^- J , length: a — |-, width: dx, area: dA = ( a — |- 1 
mass: dm = 6 dA = 6 (a - |) dx. Thus, M s = / y dm = f_f\^ \ (a + g) (a. - g) 8 dx 



dx. 



SOp- 
80-16^1 



O 2 -*)*=! [* - *] :, : - 2 • i [ A - *] 7 = s (^ - *W) 



2 \/P» 



2a 2 6 v /pa 






8a-> 



2^S 



^;M=/dm=* / (a-g) 

-2^/jS 



dx 



6 ax - 

8a<5./pa 



^-'■'h-C-"!^^ 



) = 4atf ^ (l -±)= 4a6^ (%<*) 



. Soy 



M 



/8aVpaN / 3 N 

V 5 J \m^i] 



3 



a, as claimed. 



41. Since the density is constant, its value will not affect our answers, so we can set S = 1. 

A generalization of Example 6 yields M, = J y dm = J a 2 sin dd = a 2 [— cos 3k!° 



cos (| + a) + cos (| - a)] = a 2 (sin a + sin a) = 2a 2 sin a; M = J dm = J _ "a d0 = alOfJ^l 



[(I 



*)] = 2aa. Thus, y 



Mi 
M 



2a 2 



sin a a sin a 



2aa 



Now s = a(2a) and a sin a 



c = 2a sin a. Then y = 2a" = If > as claimed. 



42. (a) First, we note that y = (distance from origin to AB) + d 



a cos a + d 



a(sin a — a cos a) 



Moreover, h = a — a cos a 




a(sin a — a cos a) sin a — a cos a 

a(a — a cos a) a — a cos a 



The graphs below suggest that 



ZOOM 

VIEW 



0.6 
0.5 
0.4 
0.3 
0.2 
0.1 



~3$2 3D~ 



/(<*) = 



sma - acosa 



5 — O.'l o!2 a 



(b) 



a 


0.2 


0.4 


0.6 


0.8 


1.0 


f(a) 


0.666222 


0.664879 


0.662615 


0.659389 


0.655145 



6.5 AREAS OF SURFACES OF REVOLUTION AND THE THEOREMS OF PAPPUS 



1. (a) |=sec 2 x^ (|) 2 = sec 4 x 

=>• S = 2n j (tan x) y 1 + sec 4 x dx 
(c) S « 3.84 



(b) 



' 








0.8 








0.6 








0.4 






y = tanjt 


0.2 











*' 0.'1 


0.'4 


6.'6 o:s 



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394 Chapter 6 Applications of Definite Integrals 



2- (a) f x =2^ (%)' = ** 

=> S = 2ir f Q x 2 a/1 +4x 2 dx 
(c) S « 53.23 



(b) 



dx 
dy 



3. (a) xy = 1 => x = ± 

=* S = 2^/ i 2 iyiT^dy 
(c) S « 5.02 



fd*\ 2 _ i 
V d y/ y 1 



(b) 





03 555 0l7 0.'8 55 f 



4. (a) | = cosy => (<|j = cos 2 y 

=4> S = 27r J (sin y) ^/l + cos 2 y dy 
(c) S « 14.42 



(b) 




U' 62 53 53 53" r 



5. (a) x 1 /2 + y i/2 = 3 ^ y= (3_ x i/2)< 
> g=2(3-xV 2 )(_I x - 1 / 2) 

,2 



(S)^ = (1-3 X - 1/2 ) 1 



=> S = 2tt J]* (3 - x 1 / 2 ) 2 ^/l + (l-3x-!/2) 2 dx 
(c) S « 63.37 



(b) 




x v J + y i/2 = 3 



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Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 

6. (a) | = l + y-i/2^ (!) 2 = (i + y -i/2) 2 (b) 

=► S = 2ir f*(y + 200 y/l + (l+ y- 1 ' 2 ) 2 dx 
(c) S « 51.33 



395 



7. (a) |=tany => (|) 2 =tan 2 y 

=> S = 2irJ (J tantdt) i/l + tan 2 y dy 
= 2tt \ I I tan t dt J sec y dy 



(c) S « 2.08 



(b) 




3 ' 3.'4 " 3.'8 ' 42 " i'&i'i x 




IIDJ'B 0:4 0:5 016 0:7 X 



(a) g = V^I^(g) =x 2 -l (b) 

=> 8 = 2^ 5 (/,Vt 2 - 1 dt) a/1 + (x 2 - 1) dx 
= 2 -/^(/ 1 X V / ^ r Tdt) xdx 



(c) S « 8.55 





1 














1.2 


y = I Jt 2 -ldt 




1 








0.8 








0.6 








0.4 








0.2 











\ vi 1:4 \:s i.'j j 


jj 





9- y=f =» g = §;s = />ry</i 



i- 



dx 



=> s = />(!) a/TT] 



TTx/5 



4 dx = T^X xdx 



Try/5 



y = 4tt\/ 5; Geometry formula: base circumference = 2-7r(2), slant height = y 4 2 + 2 2 = 2y 5 
Lateral surface area = \ (47r) I 2 y 5 ) = 47ry 5 in agreement with the integral value 



10. y = § => x = 2y => | = 2; S = J] 2™ ^1 + (|) dy = f o 2tt • 2yy / l+2 2 dy = 4^0 J fl y dy = 2^0 [y 2 ] 

= 27ry 5 • 4 = 87ry 5; Geometry formula: base circumference = 27r(4), slant height = y 4 2 + 2 2 = 2y 5 
=S> Lateral surface area = | (87r) ( 2y 5 ) = 87ry 5 in agreement with the integral value 



JiV5 Tx 2 



ii- | = i;S = />y^i + (g) L dx = j; 3 2 7 rfc±i) v / i + (i) 2 dx=^/;(x+i)dx 

= ^ [(I + 3) - (I + 1)] = ^ (4 + 2) = 37rv/5; Geometry formula: n = | + | = 1, r 2 = § + | = 2, 
slant height = y/(2 — l) 2 + (3 — l) 2 = y 5 =>■ Frustum surface area = 7r(rj + r a ) X slant height = 7r(l + 2)y 5 
= 37ry 5 in agreement with the integral value 



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396 Chapter 6 Applications of Definite Integrals 



12. y 



I => x = 2y - 1 =4> 



dx 
dy 



2; S = £l-KX Jl + (|) dy = JjWy - l)x/l + 4 dy = 2ttx/5 £(& - 1) dy 



= lir^fs [y 2 - y] J = 2tt v/5 [(4 - 2) - (1 - 1)] = 4^^; Geometry formula: n = 1, r 2 = 3, 
slant height = x/(2 — l) 2 + (3 — l) 2 = \J 5 =>■ Frustum surface area = 7r(l + 3) y 5 = Any 5 in agreement with 
the integral value 



13. 



dy 
dx 

U 

X = 



x? _^ (*£\ — X 4 . c _ f~27n 
3 =* ^dxj - 9 =^ S - Jo 9 

1 + 7 => du = | x 3 dx =>• | du = y dx; 



1 + f dx; 



H / 125 

3 V 27 



u= l,x = 2 

>25/9 
1 
1 



251 



u 1 / 2 • i du 



I \\ u 3 / 2 



25/9 



7[ ( 1 25-27 ' 
3 V 27 / 



2 L3 

98tt 

" 81 




14. 



dy _ 1 



x -l/2 



(*)' 



4x 



5<lx 



27r X" /4 \/^i dx = 27T [f (x 



4JT 

3 



15 

4 



f (8-1) 



1x3/2 

4J 

28- 
3 



1x3/2 

4) 



ix 3/21 15/4 
?J J 3/4 

f r (|) 3 



- 1 



y 

1.94- ■ 

0.87- ■ 



y=Vx_ 



0.75 



—i X 

3.75 



15- t 



1 (2 - 2x) 

2 \/2x - x 2 



1-x 



\/2x^ 



(£) : 



(i-xy 

2x — x 2 



=► s = X|;W2x-x 2 A /i 

= 2tt r 1J v / 2x-x 2 V2x ~ x ^± 

J 0.5 v x/2x 

= 2 7 r/ o ' 5 5 dx = 27r[x]i;5 = 2tt 



2 \/2x — x 2 4- 1 — 2x + x 2 j 



(1 - x) 2 
2x — x 2 

X^ 



dx 



y=V2x-x 2 



— 1 1 1 x 

.5 1 1.5 



dy _ 1 _, ( dy \ 
' dx 2v / xTT \ dx / 


1 

~ 4(x+l) 






=> S = J^ lllyjx + 1 


V 1 + 4(x 1 + l) dx 




= 27rj;Y(x+l) + i 


dx = 27Tj r | 5 A /xT 


f dx 




=2*[w+\r\\ 


= f [(5 + I) 3/2 - 


"(1 + 


D 3/2 


= 4 f[(W /2 -af 


2] 4tt /5 3 3 3 \ 
J — 3 ^23 23 J 






= |(125-27) = ^ 


49-rr 
3 







y 

2.4 ■ ■ 

1.4 -■ 



y=Vx+T 



17. 



dx ._ y 2 



dy. - (I) =y 4 ^S = X^^AT7dy; 

[u = 1 + y 4 =>- du = 4y 3 dy =>■ 1 du = y 3 dy; y = 
=> a = 1, y = 1 => u = 2] -» S = J^tt (i) u 1 / 2 (i du) 
= l/>du=l[lu 3 / 2 ]; = |(xA-l) 




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Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 397 



18. x = (i y 3 / 2 - y 1 / 2 ) < 0, when 1 < y < 3. To get positive 
area, we take x = — (4 y 3 / 2 — y 1 / 2 ) 



X= (ly^2_ y1 /2) 



l (y l/2_ y -l/ 2) ^ (|) = l(y_ 2 + y-l) 

S = - J> (I y 3 / 2 - y 1 / 2 ) ^1 + j (y - 2 + y-i) dy 



„3/2_ v l/2\ /I 




-2tt 



4 (y + 2 + y-i)dy 

/ i 3 (i^_ y i/2)^^ dy = _ ff j;V/aQy-l)(yi/» + ^)dy = -7r/ l S (iy-l)(y+l)dy 

— [(f " ! - 3) - (| - | - 1)] = — (-3 - i + | + 1) 



-|(-18-l+3) = i|2 



19. £ - 

= 4-7T 



15/4 




= 4^ => S = / Q 27r-2V4^y0T^dy = 47r/ o ^(4 - y) + 1 dy 

X' 5/4 v 7 ^ dy = -4tt [f (5 - y) 3 / 2 ] J 5/4 = - & [(5 - ¥) 3/2 - 5 3 / 2 ] = ■ - £ [(f) 3 / 2 _ 5 3 / 2 



^[(5-^) 3/2 -5 3 / 2 ]=-f [ 



55; (Sa/s - 5v_ 

3 \ J V J g 



40^-5^ ^ _ 35tt75 



\ _ 8tt A 40^5-575 \ 



on — — - 

ZU - dy - y2>^T 



(I) =2^1 =* S = J^v^T ^1 + ^T dy = 2./ v y (2y - 1) + 1 dy 

2 -X>y 1/2 dy = ^ [§ y 3 ^ 2 ] \ /8 = *£ [i 3/2 - (if 2 } = 4 -¥ - $) 



21. ds= v/dx 2 + dy 2 



y 3 -i) +!dy 



4y :; 



^(y° - I + w) + i dy = ^(y e + | + i^) dy 



(y 3 + 40 2 dy = (y 3 + fr) dy; S = J^Try ds = 2^'y (y 3 + ^) dy = 2^(y 4 + \ y- 2 ) dy 



2- [? " hi x =2- [(f - |) -(*-*)] =2.(^ + |) = 1(8-31+5)-- 



20 



22. y = \ (x 2 + 2) 3/2 =>• dy = xyV + 2 dx =>■ ds = a/I + (2x 2 + x 4 ) dx => S = 2tt/ o x v / 1+2x 2 + x 4 dx 
= 2tt/ o V/5 x v /(x 2 +1) 2 dx = 2 7 r/ o V/5 x (x 2 + 1) dx = 27r/ o "% 3 + x) dx = 2n [£ + f 1 2 = 2tt (| + §) = 4tt 



23. y = x/a 2 - x 2 => g = I (a 2 - x 2 )~ 1/2 (-2x) 



7* 



(*)' 



(a 2 - x 2 ) 



!> S = 2 7 rJ^ v / a 2 -x 2 jT+ (j^jy dx = 2irf^y/(a 2 - x 2 ) + x 2 dx = 2tt J\ dx = 27ra[x]l a 
2?ra[a - (-a)] = (27ra)(2a) = 4?ra 2 



24. y 



dy 

dx h 



h ^ \dx) 



2jrr 
h 



25. y = cos x — - t - — sin x 



dy 
dx 



= £=► 


S = 


2tt r 

Jo 


'^V 1+ P 


'h 2 + r 2 [ 


X 2 ] 

2 Jo 


27rr 
~ h 2 


x/h 2 + r 2 (^ 


(D' = 


sin 2 


X => 


f.jr/2 

S = 2tt , 

J-jr/2 



+ H 



dx 



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398 Chapter 6 Applications of Definite Integrals 

26. y = (1 - x*/3) 3/2 => S = | (1 - x 2 / 3 ) 1/2 (- 1 x-V3) = _ (±^ 2 



dx 2 

.2/3^ 3 / 2 



xV3 



(*)' 



1-x 2 / 3 _ J_ 
x 2/3 — x 2/3 



> S = 2/ o 2tt (1 - x 2 /3) 3/2 ^1 + (-^ - l) dx = 4irf o (l - x 2 / 3 ) 3/2 ^V* dx 
47r/ ' (1 - x 2 / 3 ) 3/2 x-V3 dx; [u = 1 - x 2 / 3 => du = - § x" 1 ^ dx - 
= 0=>u=l,x=l =^ u = 0] -► S = 4tt J" u 3 / 2 (- | du) = -6tt [| u 5 / 2 ] ° = -6tt (0 



\ du = x 1/<3 dx; 



2\ _ 12tt 
5/ — 5 



27. The area of the surface of one wok is S = J 2nx J 1 + ( ^ J dy. Now, x 2 + y 2 = 16 2 =4> x = ^/W~^f 

= *£? ; s = £2-v / T6 2 ^7V 1 + T6^ d y = 2-£>(i6 2 -y 2 ) + y 2 dy 



dy - ^/162-y 2 "^ 

= 2tt I 16 dy = 327r • 9 = 2887T « 904.78 cm 2 . The enamel needed to cover one surface of one wok is 

V = S • 0.5 mm = S • 0.05 cm = (904.78)(0.05) cm 3 = 45.24 cm 3 . For 5000 woks, we need 
5000 • V = 5000 • 45.24 cm 3 = (5)(45.24)L = 226.2L => 226.2 liters of each color are needed. 



28.y = v / r^ 2 "^ g = - \ -fc 



(I) =^;s=2./; Vr^^yrr^ 



Vi' 2 — x 2 vr 2 — x 2 

Xa+h _ /»a+h 

y (r 2 — x 2 ) + x 2 dx = 27rr I dx = 27rrh, which is independent of a. 



dx 



29. y = \/R 2 - x 2 => 



dy _ _ l 2x . -x 

dx " 2 y R 2 _ x 2 — ,/ R 2 - x 2 

a+h 



(I) =p^;s = 2./; v^ 2 " 



x 2 ./l + p^dx 



= 2nf* v/(R 2 -x 2 )+x 2 dx = 2ttR J & dx = 2?rRh 

30. (a) x 2 + y 2 = 45 2 => 



^45^7 



2 _v dx 



dy v /452_ y 2 



(I) 1 



452 _ y 2 



S = f 45 2tt a/45 2 
J -22.5 v 



y\n 



-22.5 v J V ' 45 2 -y 2 

= (2?r)(45)(67.5) = 6075tt square feet 
(b) 19,085 square feet 



dy = 2n Xl 5 v / (45 2 -y 2 )+y 2 dy = 2tt ■ 45^/y 



31. y = x => (|)=1 => (|) =1 => S = 2^/* |x| y/l + 1 dx = 2tt f^(-x)y/2 dx + 2wf\y/2 dx 
= -2\/27r[f] + 2^/277 [f] = -2\/27r(0- i) + 2 v / 2tt(2-0) = 5^ 



32- I = f 



33. 



f |Q = ^ ^ by symmetry of the graph that S = 2 J_°^ 2tt (- f ) ^/l + f dx; [u = 1 + f 
=> du = \ x 3 dx => - 1 du = - f dx; x = -^3 ^u = 2, x = 0^u=lj -> S = 47r/ 2 ' u 1 / 2 (- \) du 
= — 7T J u 1 / 2 du = — 7T [| u 3//2 ] = — 7T ( | — | \/8 = y ( \/8 — 1 J . If the absolute value bars are dropped the 

integral for S = J _ 27rf(x) ds will equal zero since I _ 27r I ^- ) J 1 + ^ dx is the integral of an odd function 
over the symmetric interval — \J 3 < x < y\3. 

^ = -sintand| = cost => ^ (^f) 2 + (f )' = V(- sin t) 2 + (cos t) 2 = 1 => S = / 27ry ds 
= f " 2n(2 + sin t)(l) dt = 2tt [2t - cos t] jf = 27r[(47r - 1) - (0 - 1)] = 8tt 2 



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Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 

34. | = & and | = r^ =» a/(|) 2 + (|) 2 = a/H^ = ^ => S = / 2™ ds 



399 



J. 2 ^ 



2tsm j£+i 



dt 



4£ 



-V5 



/ t^t 2 + 1 dt; [u = t 2 + 1 => du = 2t dt; t = => u = 1, 



[t = a/3 => u = 4J -» J/f A/udu = [f u 3 / 2 ] J 



:s.. 

9 



v/3 



3/21 J 2+1 



dt is an improper integral but lim f(t) exists and is equal to 0, where 



Note: I 27r(ft 

f(t) = 2tt (I t 3 / 2 ) v/^ 1 . Thus the discontinuity is removable: define F(t) = f(t) for t > and F(0) = 

=* Jo F « dt 



2S:. 

9 



,K 'and & 
'V5 



35. ^ = land^= . 



. dtJ 



(t)' 



t 2 + 2A/2t+3 =^> S = /2?rxds 



= /_! 2tt ft + a/2) yt 2 + 2y2t + 3 dt; [u = t 2 + 2-^2 1 + 3 =>• du = ^2t + 2a/2J dt; t = - \/2 => u = 1, 

t = a/2 => u = 9J -» /Va/u du = [§ 7m 3 / 2 ] j = f (27 - 1) 



52^ 
3 



36 — 

JU - dt 



a(l — cos t) and $ = a sin t 



y(|) 2 +(f) 2 =^[a(l-cost)] 2 + (asint) 2 
V a 2 — 2 a 2 cos t + a 2 cos 2 t + a 2 sin 2 t = \J 2a 2 — 2a 2 cos t = ay 2 v 1 — cos t =>- S = J 27ry ds 
J *2tt a(l - cos t) • a a/2 a/ 1 -cost dt = 2 a/2 tt a 2 J^l - cos t) 3/2 dt 



37. | = 2and| = l =► ^f) 2 + (f) 2 = ^^ = ^ => S = J 2ny ds = /J 2^(t + Dy^dt 

/- r . -I 1 /- /- /- /- 

= 2-7TV5 I + t = 37rV5. Check: slant height is V 5 => Area is 7r(l + 2)a/5 = 3ir a/5 . 



38. £=hand£ 



dx\ 2 
. dt J 



) = A/h 2 + r 2 => S = / 2?ry ds = J u 2 7 rrtA/h 2 ~+r 2 d 
27rrA/h 2 + r 2 J t dt = 27rrA/h 2 + r 2 l~|| = 7rrA/h 2 + r 2 . Check: slant height is \/h 2 + r 



Area is 



rv/h 2 " 



y=/W 



39. (a) An equation of the tangent line segment is 
(see figure) y = f(m k ) + f'(m k )(x - m k ). 
When x = x k _i we have 
ri = f(m k ) + f'(m k )(x t _, - m k ) 

= f(m k ) + f'(m k ) (- Af) = f(m k ) - f'(m k ) ^ ; 
when x = x k we have 
T2 = f(m k ) + f'(m k )(x i - m k ) 

= f(m k ) + f '(m k ) %* ; 

(b) L 2 = (Ax k ) 2 + (r 2 - n) 2 

= (Ax k ) 2 + [f'(m k ) *? - (-f'(m k ) ^)] 2 

= (Ax k ) 2 + [f'(m k )Ax k ] 2 => L k = A/(Ax k ) 2 + [f'(m k )Ax k ] 2 , as claimed 

(c) From geometry it is a fact that the lateral surface area of the frustum obtained by revolving the tangent 




line segment about the x-axis is given by AS k = 7r(ri + r2)Lk = 7r[2f(m k )] y (Ax k )" + [f'(m k )Ax k ] 2 
using parts (a) and (b) above. Thus, AS k = 27rf(m k ) a/1 + [f'(m k )] 2 Ax k 



Copyright (c) 2006 Pearson Education, Inc 




400 Chapter 6 Applications of Definite Integrals 

(d) S = n Km o £ AS t = B Bm ± 2rf(m k ) ^1 + [f (m t )] 2 Ax k = £ 2rf(x) v/l + [f (x)] 2 dx 



40. S = J>f(x) dx = I 3 2n- -fc dx = -fc [x 2 ]f = ^ = V^tt 



^ UA " n/3 



75 



41. The centroid of the square is located at (2, 2). The volume is V = (2tt) (y) (A) = (2tt)(2)(8) = 32tt and the 
surface area is S = (2tt) (y) (L) = (27r)(2) I 4y8 ) = 32\/27r (where \J 8 is the length of a side). 



42. The midpoint of the hypotenuse of the triangle is ( | , 3) 
=4- y = 2x is an equation of the median =4> the line 
y = 2x contains the centroid. The point ( | , 3) is 

-¥- units from the origin =>■ the x-coordinate of the 



centroid solves the equation y (x - 


-|) 2 + (2x- 


-3) 2 


= 4 => (x 2 -3x+|)+(4x 2 - 


- 12x + 9) = 


5 
" 4 




=*> 5x 2 - 15x + 9 = -1 

=4> x 2 — 3x + 2 = (x — 2)(x — 1) = => x = 1 since the centroid must lie inside the triangle => y = 2. By the 
Theorem of Pappus, the volume is V = (distance traveled by the centroid)(area of the region) = 2ir (5 — x) [| (3)(6)] 
= (2tt)(4)(9) = 72tt 



43. The centroid is located at (2,0) 



(2?r) (x) (A) = (27r)(2)(7r) = 4tt 2 



44. We create the cone by revolving the triangle with vertices 
(0, 0), (h, r) and (h, 0) about the x-axis (see the accompanying 
figure). Thus, the cone has height h and base radius r. By 
Theorem of Pappus, the lateral surface area swept out by the 
hypotenuse L is given by S = 2-7ryL = 27T (§) y h 2 + r 2 

= 7rrv r 2 + h 2 . To calculate the volume we need the position 
of the centroid of the triangle. From the diagram we see that 



(h.r) 




the centroid lies on the line y = jr x. The x-coordinate of the centroid solves the equation \f (x — h) 



\^i^^)^~^)^^^ 



+ 4h 2 ) 







2h or 4h 
3 UI 3 



, since the centroid must lie 



inside the triangle 



2h 



\. By the Theorem of Pappus, V = [2tt fj )] [\ hr) = 1 7rr 2 h. 



45. S = 27ry L =>• 47ra 2 = (27ry) (7ra) =>- y = ^, and by symmetry x = 

46. S = 27rpL => [2tt (a - f )] (vra) = 27ra 2 (7r - 2) 



47. V = 27ryA => | ?rab 2 = (27ry) 



48. V = 2^pA => V = [2tt (a + £)] (^ ) 



4b 



y = 5^ and by symmetry x = 



-a 3 (3-+4) 
3 



49. V = 2irp A = (27r)(area of the region) • (distance from the centroid to the line y = x — a). We must find the 
distance from (0, y\ to y = x — a. The line containing the centroid and perpendicular to y = x — a has slope 



— 1 and contains the point (0, |^) . This line is y = — x + ||. The intersection of y = x — a and y = — x + |Ms 



the point 



f4a+3a7r 4a — 3a^^^ 



(lK 



6n 



Thus, the distance from the centroid to the line y 



x — a is 



Copffigl (c| 1 Pearson Educate to, publishing as Pearson Addison-Wesle 



Section 6.6 Work 401 



'4a + 3a7T\ 2 , /4a _ 4a , 3a7r\ 2 _ y/2 (4a + 3a;r) v _ n , / y|(4a + 3a7r) \ /jraA _ 72 ;ra 3 (4 + 3?r) 

, 6tt J + V3tt 6ir + 6tt J — 6jr ^ V — (Z7T; ^ 6?r y^ 2 y_ fi 



50. The line perpendicular to y = x — a and passing through the centroid (0, — ) has equation y = — x + — . The 



intersection of the two perpendicular lines occurs when x — a = — x + — =>• x = 2a 2 l ^ r a7T 



y= 2a_^ Thus 



the distance from the centroid to the line y = x - a is \j ( ^Mp - 0) 2 + ( ^^ - | ) 2 = ^^ 



Therefore, by the Theorem of Pappus the surface area is S = 2ir 






(vra) = V27ra 2 (2 + 7r). 



5 1 . From Example 4 and Pappus's Theorem for Volumes we have the moment about the x-axis is M x = y M 



»)(¥) 



2ir: 
3 



6.6 WORK 



1 . The force required to stretch the spring from its natural length of 2 m to a length of 5 m is F(x) = kx. The 



work done by F is W = I F(x) dx = k I x dx = | [x 2 ] 
=^ k = 400N/m 



9k 

2 • 



This work is equal to 1 800 J 



1800 



^f = 200 lb/in. 



2. (a) We find the force constant from Hooke's Law: F = kx =>• k = - =>• k 

(b) The work done to stretch the spring 2 inches beyond its natural length is W = I kx dx 
= 200 J q x dx = 200 |*f 1 = 200(2 - 0) = 400 in • lb = 33.3 ft • lb 



(c) We substitute F = 1600 into the equation F = 200x to find 1600 = 200x 



in. 



3. We find the force constant from Hooke's law: F = kx. A force of 2 N stretches the spring to 0.02 m 

^> 2 = k - (0.02) => k = 100 Si. The force of 4 N will stretch the rubber band y m, where F = ky => y = 

X0.04 
kx dx 



4N 
100 g 



r-0.04 

100 x 

Jo 



dx = 100 5 



0.04 




(100X0.04) 2 

2 



0.08 J 



4. We find the force constant from Hooke's law: F = kx 



I => k=2<> => k : 



90 — . The work done to 



stretch the spring 5 m beyond its natural length is W = I kx dx = 90 I x dx = 90 - 



(90) 



'25^ 



1125 J 



5. (a) We find the spring's constant from Hooke's law: F = kx=>k=| = 2LZ11 = 21£li => k = 7238 Pj 

n0.5 n0.5 

(b) The work done to compress the assembly the first half inch is W = I kx dx = 7238 I x dx 

= 7238 [f 1 = (7238) ^ = < 7238 >< - 25 > « 905 in • lb. The work done to compress the assembly the 



second half inch is: W = J^ kx dx = 7238 J^ x dx = 7238 [f 1 = ^ [1 - (0.5) 



(7238)(0.75) 
2 



2714 in -lb 



6. First, we find the force constant from Hooke's law: F = kx => k = - = 41^- = 16 • 150 = 2,400 *. If someone 

(re) 

compresses the scale x = I in, he/she must weigh F = kx = 2,400 (|) = 300 lb. The work done to compress the 

f.l/8 

scale this far is W = J o kx dx = 2400 



1/8 



§^ = 18.75 lb- in. = ff ft - lb 



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402 Chapter 6 Applications of Definite Integrals 



7. The force required to haul up the rope is equal to the rope's weight, which varies steadily and is proportional to 
x, the length of the rope still hanging: F(x) = 0.624x. The work done is: W = I F(x) dx = I 0.624x dx 

= 0.624 \4\ =780 J 



8. The weight of sand decreases steadily by 72 lb over the 1 8 ft, at 4 lb/ft. So the weight of sand when the bag is x ft off the 
ground is F(x) = 144 - 4x. The work done is: W = J F(x) dx = j o (144 - 4x)dx = [144x - 2x 2 ] f = 1944 ft • lb 



9. The force required to lift the cable is equal to the weight of the cable paid out: F(x) = (4.5)(180 — x) where x 

/>180 nlSO 

is the position of the car off the first floor. The work done is: W = / F(x) dx = 4.5 I (180 — x) dx 



r 2i 180 i 

[iSOx-V] =4.5( 



4.5 180x 



4.5 180 2 



180 2 



4.5-180 2 



72,900 ft • lb 



10. Since the force is acting toward the origin, it acts opposite to the positive x-direction. Thus F(x) 



t. The 



work done is W 



r 



-^dx 



kf^dx= k [i]:=k(i 



k(a - b) 
ab 



11. The force against the piston is F = pA. If V = Ax, where x is the height of the cylinder, then dV = A dx 



Work = J F dx = J pA dx = J p 



dV. 



12. pV = c, a constant => p 

Thus W = £ 109,350V- 14 dV = [- ™>] 



'243 

(109,350)(5) 
(0.4)(36) 



cV" 1 - 4 . If Vi = 243 in J and pi = 50 lb/in d , then c = (50)(243) 

32 

243 



109,3501b. 



109,350 ( 1 
v 32 - 4 



0.4 



243°- 4 ; 



109.350 (1 
0.4 U 



-37,968.75 in • lb. Note that when a system is compressed, the work done by the system is negative. 



13. Let r = the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate, 

the amount of water in the bucket is proportional to (20 — x), the distance the bucket is being raised. The leakage rate of 

the water is 0.8 lb/ft raised and the weight of the water in the bucket is F = 0.8(20 — x). So: 

r 20 r 2i 20 

W = J o 0.8(20 - x) dx = 0.8 20x - f I = 160 ft • lb. 



14. Let r = the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate, 

the amount of water in the bucket is proportional to (20 — x), the distance the bucket is being raised. The leakage rate of 

the water is 2 lb/ft raised and the weight of the water in the bucket is F = 2(20 — x). So: 

r 20 r 21 20 

W = J o 2(20 - x) dx = 2 20x - \\ = 400 ft • lb. 

Note that since the force in Exercise 14 is 2.5 times the force in Exercise 13 at each elevation, the total work is also 2.5 
times as great. 



Copyright (c) 2006 Pearson Education 




Section 6.6 Work 403 



G *ou od 10ft 



15. We will use the coordinate system given. 

(a) The typical slab between the planes at y and y + Ay has 
a volume of AV = (10)(12) Ay = 120 Ay ft 3 . The force 
F required to lift the slab is equal to its weight: 
F = 62.4 AV = 62.4 • 120 Ay lb. The distance through 
which F must act is about y ft, so the work done lifting 
the slab is about AW = force x distance 
= 62.4 - 120 • y • Ay ft • lb. The work it takes to lift all 

20 

the water is approximately WrsJ] AW 



20 

= J2 62.4 • 120y • Ay ft • lb. This is a Riemann sum for 



the function 62.4 • 120y over the interval < y < 20. The work of pumping the tank empty is the limit of these sums: 




n20 r 21 20 

W = J o 62.4 • 120y dy = (62.4)(120) I \ I = (62.4)(120) (*f) = (62.4)(120)(200) 



-hp motor is t 



w 

250 fHb 



1,497,600 ft-lb 
250 Mb 



1,497,600 ft -lb 
5990.4 sec 



(b) The time t it takes to empty the full tank with ( -^ 

= 1 .664 hr =4> t « 1 hr and 40 min 

(c) Following all the steps of part (a), we find that the work it takes to lower the water level 10 ft is 

W = J o 62.4 • 120y dy = (62.4)(120) NU = (62.4)(120) (±f) = 374,400 ft • lb and the time is t 

= 1497.6 sec = 0.416 hr w 25 min 

(d) In a location where water weighs 62.26 — • 



250 ^ 



ft 3 ■ 



a) W = (62.26)(24,000) = 1,494,240 ft • lb. 



b)t 



1,494,240 
250 



5976.96 sec w 1.660 hr 



In a location where water weighs 62.59 



IV" 



a) W = (62.59)(24,000) = 1,502,160 ft - lb 



b)t 



1,502,160 
250 



6008.64 sec w 1.669 hr 



1 hr and 40 min 



1 hr and 40.1 min 



16. We will use the coordinate system given. "Trr^^ .Ground level 

(a) The typical slab between the planes at y and y + Ay has 
a volume of AV = (20)(12) Ay = 240 Ay ft 3 . The force 
F required to lift the slab is equal to its weight: 
F = 62.4 AV = 62.4 • 240 Ay lb. The distance through 
which F must act is about y ft, so the work done lifting 
the slab is about AW = force x distance 

20 

= 62.4 • 240 • y • Ay ft • lb. The work it takes to lift all the water is approximately W w J2 AW 

10 

20 

= ^2 62.4 • 240y • Ay ft • lb. This is a Riemann sum for the function 62.4 • 240y over the interval 

10 

/>20 

10 < y < 20. The work it takes to empty the cistern is the limit of these sums: W = 1 62.4 • 240y dy 




2l) 



(b) t 



(62.4)(240) I \ I = (62.4)(240)(200 - 50) = (62.4)(240)(150) = 2,246,400 ft • lb 

2,246,400 ft-lb 
275 



W 



275 ! 



8168.73 sec w 2.27 hours w 2 hr and 16.1 min 



(c) Following all the steps of part (a), we find that the work it takes to empty the tank halfway is 
W = J^62.4 • 240y dy = (62.4)(240) [y| n = (62.4)(240) (^ - ±f) = (62.4)(240) ' 125 ^ 



936,000 ft. 



Then the time is t 



\Y 



275 ! 



936,000 

275 



. J 10 

3403.64 sec 



56.7 min 



Copyright (c) 2006 Pearson Education 




404 Chapter 6 Applications of Definite Integrals 



(d) In a location where water weighs 62.26 — • 



A» ■ 



a) W = (62.26)(240)(150) = 2,241,360 ft - lb. 



b)t 



2,241,360 

275 



8150.40 sec = 2.264 hours « 2 hr and 15.8 min 



933,900 



c) W = (62.26)(240) (±f ) = 933,900 ft • lb; t - , 7 , 
In a location where water weighs 62.59 p 

a) W = (62.59)(240)(150) = 2,253,240 ft - lb. 

b) t = 2 ' 25 2 jf° = 8193.60 sec = 2.276 hours ra 2 hr and 16.56 min 

c) W = (62.59)(240) (^) = 938,850 ft • lb; t = ^^ « 3414 sec 



3396 sec « 0.94 hours rj 56.6 min 



0.95 hours ss 56.9 min 



17. The slab is a disk of area 7rx 



thickness Ay, and height below the top of the tank (10 — y). So the work to pump 



the oil in this slab, AW, is 57(10 — y)7r(|) . The work to pump all the oil to the top of the tank is 
W = X'^CLOy 2 - y 3 )dy = ^ U^f- - £1 = 11,875tt ft • lb » 37,306 ft • lb. 



18. Each slab of oil is to be pumped to a height of 14 ft. So the work to pump a slab is (14 — y)(7r) (|) and since the tank is 



half full and the volume of the original cone is V = ^7rr 2 h = i7r(5 2 )(10) 



with half the volume the cone is filled to a height y. 



■j 'in. 

G 



1 v 2 

3^4y^y 



^ ft 3 , half the volume 

7500, 



250?r fi-3 



' 500 ft. SoW 



/; 



ft\ and 



[ (14y 2 -y 3 ) dy 



577T |~ 14y 3 
4 3 



: r.( ii ) 



60,042 ft • lb. 



19. The typical slab between the planes at y and and y + Ay has a volume of AV = 7r(radius) 2 (thickness) 



'20\ 2 



Ay 



= 7T • 100 Ay ft 3 . The force F required to lift the slab is equal to its weight: F = 51.2 AV = 51.2 • 1007T Ay lb 
=^> F = 51207T Ay lb. The distance through which F must act is about (30 — y) ft. The work it takes to lift all the 

30 30 

kerosene is approximately W w Y AW = Y 51207r(30 — y) Ay ft • lb which is a Riemann sum. The work to pump the 





tank dry is the limit of these sums: W= I 51207r(30 — y) dy = 51207T 30y 
w 7,238,229.48 ft • lb 



::ii 



5120^(250) 



(5120)(450tt) 



20. (Alternate Solution) Each method must pump all of the water the 15 ft to the base of the tank. Pumping to the rim requires 
all the water to be pumped an additional 6 feet. Pumping into the bottom requires that the water be pumped an average of 3 
additional feet. Thus pumping through the valve requires y 3 ft(47r)6 ft 3 (62.4 lb/ft 3 ) « 14,1 15 ft • lb less work and thus 
less time. 



21. (a) Follow all the steps of Example 5 but make the substitution of 64.5 p for 57 p. Then, 



W 



X 



o S^s: (10 - y)y 2 dy 



64.5tt ioy 
4 3 



64.5 

4 



E ( 'lOJ 

V 3 



' 64.5 
v 4 



-) (8 3 ) 



-2 



64.5;r-i 
3 



21.5tt-8 3 « 34,582.65 ft • lb 



(b) Exactly as done in Example 5 but change the distance through which F acts to distance w (13 — y) ft. 



Then W = p Q Z* (13 - y)y 2 dy = ^ [ 
= (19tt) (8 2 ) (7)(2) w 53.482.5 ft - lb 



13y 3 
3 



r] = 5J r{ 1J f~^) = ( 5J r)(S 3 )( l i-2) 



57tt-& j -7 
3-4 



22. The typical slab between the planes of y and y+Ay has a volume of about AV = 7r(radius) 2 (thickness) 
= 7T (^/y) 2 Ay = xy Ay m 3 . The force F(y) is equal to the slab's weight: F(y) = 10,000 ^ • AV 
= 7iT0,000y Ay N. The height of the tank is 4 2 = 16 m. The distance through which F(y) must act to lift 
the slab to the level of the top of the tank is about (16 — y) m, so the work done lifting the slab is about 

Copyright (c| 1 Pearson Etation, Inc., publishing as Pearson Addison-Wesle 



Section 6.6 Work 405 



AW = 10,0007ry(16 — y) Ay N • m. The work done lifting all the slabs from y = 0toy= 16 to the top is 

16 

approximately W w Yl 10,0007ry(16 — y)Ay. Taking the limit of these Riemann sums, we get 
o 

16 



W = J 10,0007ry(16 - y) dy = 10,000tt/ (16y - y 2 ) dy = 10,000tt 
21,446,605.9 J 



16y 2 
2 



10,000tt ( 



16 3 16 3 



1 0.000- tt-16 3 
6 



23. The typical slab between the planes at y and y+Ay has a volume of about AV = 7r(radius) 2 (thickness) 



= 7r (a/25 — y 2 ) Ay m 3 . The force F(y) required to lift this slab is equal to its weight: F(y) = 9800 • AV 

= 9800tt (a/25 -y 2 ) 2 Ay = 9800?r (25 - y 2 ) Ay N. The distance through which F(y) must act to lift the 
slab to the level of 4 m above the top of the reservoir is about (4 — y) m, so the work done is approximately 
AW « 98007T (25 - y 2 ) (4 - y) Ay N • m. The work done lifting all the slabs from y = -5 m to y = m is 



approximately W rs ]T] 98007T (25 — y 2 ) (4 — y) Ay N • m. Taking the limit of these Riemann sums, we get 

-5 

W = J_° 5 98007r (25 - y 2 ) (4 - y) dy = 9800tt /_° 5 (100 - 25y - 4y 2 + y 3 ) dy = 9800tt [lOOj/ - f y 2 - \ y 3 
= -9800tt (-500 - ^ + § • 125 + ^f) * 15,073,099.75 J 



24. The typical slab between the planes at y and y+Ay has a volume of about AV = 7r(radius) (thickness) 

= n (a/100 -y 2 ) 2 Ay = tt (100 - y 2 ) Ay ft 3 . The force is F(y) = ^ • AV = 56?r (100 - y 2 ) Ay lb. The 
distance through which F(y) must act to lift the slab to the level of 2 ft above the top of the tank is about 
(12 - y) ft, so the work done is AW « 567T (100 - y 2 ) (12 - y) Ay lb - ft. The work done lifting all the slabs 

10 

from y = ft to y = 10 ft is approximately W « Yl 567r ( 100 — y 2 ) (12 — y) Ay lb • ft. Taking the limit of these 



Riemann sums, we get W = f 56tt (100 - y 2 ) (12 - y) dy = 56ttJ o (100 - y 2 ) (12 - y) dy 
= 567r/ o 10 (1200 - lOOy - 12y 2 + y 3 ) dy = 56tt [l200y 



1 00y- 

2 



nf 

3 



10 







56tt (12,000 



10,000 



4 - 1000 + ISfiQO) 



(56tt)(12-5-4- 



. .„„.,, . , ,,._ .. , .,, (1000) w 967,611 ft -lb. 

It would cost (0.5)(967,61 1) = 483,8050 = $4838.05. Yes, you can afford to hire the firm. 



25. F = m 



dv 



mv p- by the chain rule 



W 



/: 



mv fr dx = m 



/: 



dv i 



dx = m [i v 2 (x 



\ m [v 2 (x2) — v 2 (xi)] = | mv| — \ mv 2 , as claimed. 



26. weight = 2 oz = ^ lb; mass = "^ = 55 = 555 slu § s ; W = (1/ 



27. 90mph 

W 



90 mi 1 hr 



1 min 5280 ft 



1 hr 60 min 60 sec 1 mi 

l)(iiii)( 132ft/sec ) 2 « 85 - lft - lb 



132 ft/sec; m = |^ 



(255 slugs) (160 ft/sec) 2 w 50 ft- lb 
°-^ slugs; 



28. weight = 1.6oz = 0.1 lb 



0.1 lb 

32 ft/sec 2 



L slugs; W 



320 



3I0 slugs) (280 ft/sec) 2 = 122.5 ft • lb 



29. weight = 2 oz 

W 



I lb 



m = A slugs 



1 slugs; 124 mph = (1 f fi 4 ,^ 0) w 181.87 ft/sec; 



25 ft 



(60)(60) 



i) (^ slugs) (181.87 ft/sec) 2 « 64.6 ft • lb 



30. weight = 14.5 oz 



14=5 lb 

16 1U 



14.5 

(16)(32) 



slugs; W = (i) ((i^) slugs j (88 ft/sec) 2 w 109.7 ft ■ 



lb 



31. weight = 6.5 oz = j| lb =$> m 



6.5 

(16)(32) 



slugs; W = (1) ((1H2) slu S s ) ( 132 ft/sec) 2 « 110.6 ft • j 

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406 Chapter 6 Applications of Definite Integrals 



32. F = (18 lb/ft)x => W = J* 18x dx = [9x 2 ] J /6 = \ ft - 



lb. Now W = \ mv 2 - 1 mv 2 , where W = 5 ft- lb, 



32 ~~ 256 



slugs and vi = ft/sec. Thus, 4 ft • lb 



,2J V256 



slugs) v 2 =>■ v = 8 V 2 ft/sec. With v = 



at the top of the bearing's path and v = 8 y 2 — 32t =>• t = ■ 2 4- sec when the bearing is at the top of its path. 



The height the bearing reaches is s = 8 y 2 1 — 16t 2 



8v / 2)(^)-(16)(^) 2 = 2ft 



att 



the bearing reaches a height of 



33. (a) From the diagram, 

r(y) = 60 - x = 60 - y^O 2 - (y - 325) 2 

for 325 < y < 375 ft. 
(b) The volume of a horizontal slice of the funnel 

is AV«7r[r(y)] 2 Ay 



60- v '50 2 -(y-325)^ 



Ay 



(c) The work required to lift the single slice of 
water is AW sa 62.4AV(375 - y) 



62.4(375 - y)7r 



60 



50 2 - (y - 325) 5 



Ay. 



The total work to pump our the funnel is W 



fSJ5 

J 325 62.4(375 - y)7r 
6.3358- 10 7 ft ■ lb. 



60 



50 2 - (y - 325) 



dy 




y-325 



34. (a) From the result in Example 6, the work to pump out the throat is 1,353,869,354 ft • lb. Therefor, the total work 
required to pump out the throat and the funnel is 1,353,869,354 + 63,358,000 = 1,417227,354 ft ■ lb. 
(b) In horsepower-hours, the work required to pump out the glory hole is ' { ^ 1( ^r =715.8. Therefore, it would take 
7 i„ 8 h h p p h = 0.7158 hours « 43 minutes. 



35. We imagine the milkshake divided into thin slabs by planes perpendicular to the y-axis at the points of a 
partition of the interval [0, 7]. The typical slab between the planes at y and y + Ay has a volume of about 



AV = 7r(radius) 2 (thickness) = 7r 



>y+VL5\ 2 



14 



Ay in 3 . The force F(y) required to lift this slab is equal to its 



weight: F(y) = | AV = y ( y+ 14 ' ) Ay oz. The distance through which F(y) must act to lift this slab to 
the level of 1 inch above the top is about (8 — y) in. The work done lifting the slab is about 



AW= ¥ 



4tt\ (y+17.5) : 



42^- (8 — y) Ay in • oz. The work done lifting all the slabs from y = to y = 7 is 



approximately W = Yl 9^52 (y + 17.5) 2 (8 — y) Ay in • oz which is a Riemann sum. The work is the limit of 



these sums as the norm of the partition goes to zero: W = I y^jp (y + 17.5) 2 (8 — y) dy 



¥w f ( 2450 - 26 - 25 y - 2? y 2 - y 3 ) 4y=fo\-4- 9y 3 - 2J f y 2 + 2451 )>< 



4tt _ Zi _ Q 73 _ 26.25 7 2 
9-142 4 y ' ' 2 ' ' 



9-14- [ 4 

2450 71 w 91.32 in • oz 



36. We fill the pipe and the tank. To find the work required to fill the tank follow Example 6 with radius = 10 ft. Then 

AV = 7T • 100 Ay ft 3 . The force required will be F = 62.4 • AV = 62.4 • 100tt Ay = 6240?r Ay lb. The distance through 
which F must act is y so the work done lifting the slab is about AWi = 62407T • y - Ay lb - ft. The work it takes to 



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Section 6.7 Fluid Pressures and Forces 407 



385 385 

lift all the water into the tank is: Wi « J2 AWi = J2 62407T - y • Ay lb • ft. Taking the limit we end up with 

360 360 

- 385 r ,1 385 „,„ 

= ^ [385 2 - 360 2 ] 

360 2 

To find the work required to fill the pipe, do as above, but take the radius to be | in = g ft. 
Then AV = n • A Ay ft 3 and F = 62.4 • AV 



Wi = J 36o 62407ry dy = 6240tt |"f j 



182,557,949 ft -lb 



^j 21 Ay. Also take different limits of summation and 



integration: W2 ~ J2 AW2 



Wo 



r 

Jo 



62.4 
36 



7ry dy 



62.4tt yf 
36 2 



:!6(i 



1 62Air ' 
v 36 . 



(¥) 



352,864 ft • lb. 



The total work is W = Wi + W 2 « 182,557,949 + 352,864 w 182,910,813 ft • lb. The time it takes to fill the 



tank and the pipe is Time = j^ 



182.910.813 
1650 



110,855 sec w 31 hr 



37. Work = i^o^MG dr = 1000M G 

J 6,370,000 r^ J 6 ; 37C 

= (1000) (5.975 - 10 24 ) (6.672 • lO" 11 ) (^ 



dr 



i/w\a*/~i r n 35.780.000 
1000 MG [- ?] 6,370,000 



35,780,000 



5.144 x 10 10 J 



38. (a) Let p be the x-coordinate of the second electron. Then r 2 = (p — l) 2 



W 



J°_F(p) dp 



1: 



(23xlQ- 29 ) 



i i/'-ii J dp ~ L /■-' j-i 



23xl0" 29 



(23 x 10" 29 ) (1 



11.5 x 10 



-29 



(b) W = Wi + W2 where Wi is the work done against the field of the first electron and W2 is the work done 
against the field of the second electron. Let p be the x-coordinate of the third electron. Then r 2 = (p — l) 2 



and 



r 2 = (p + I) 2 =* Wl = X 5 ^ dp = J 3 5 f^ dp = -23 x 10- 29 [ 



29 I 1 



(-23 x 10" 29 ) 



23 



x 10 



-29 



and W9 



-23xl0- 29 U^ 



4 " - > '2 = 

(-23 x 10- 29 ) (1 - i) 



1: 



23xl0" 29 



^1^^ 



^=^ (3 - 2) = Q x 10- 29 . Therefore 



12 



W = W x + W 2 = (f x 10- 29 ) + (| x 10- 29 ) 



f x 10- 29 



7.67 x 10 



-29 



6.7 FLUID PRESSURES AND FORCES 



1 . To find the width of the plate at a typical depth y, we first find an equation for the line of the plate's 

right-hand edge: y = x — 5. If we let x denote the width of the right-hand half of the triangle at depth y, then 
x = 5 + y and the total width is L(y) = 2x = 2(5 + y). The depth of the strip is (— y). The force exerted by the 

water against one side of the plate is therefore F = I w(— y) • L(y) dy = I 62.4 • (— y) - 2(5 + y) dy 
= 124.8 J_" 5 2 (-5y - y 2 ) dy = 124.8 [- § y 2 - \ y 3 ] ~j = 124.8 [(- f - 4 + \ ■ 8) - (- f • 25 + \ ■ 125)] 
= (124.8) (V - W- 



y 2 ) dy = 124.8 [- | y 2 

) = (124.8) {^m-- 



1684.8 lb 



2. An equation for the line of the plate's right-hand edge is y = x — 3 =S> x=y + 3. Thus the total width is 
L(y) = 2x = 2(y + 3). The depth of the strip is (2 — y). The force exerted by the water is 

F = J\w(2 - y)L(y) dy = J^62.4 • (2 - y) • 2(3 + y) dy = 124.8/^(6 - y - y 2 ) dy = 124.8 [6y - £ - ^ 

= (-124.8) (-18- |+9) = (-124.8) (-f) = 1684.81b 



3. Using the coordinate system of Exercise 4, we find the equation for the line of the plate's right-hand edge is 

y = x — 3 => x = y + 3. Thus the total width is L(y) = 2x = 2(y + 3). The depth of the strip changes to (4 — y) 

=>" F = /I w < 4 - y)L(y) dy = J_° 3 62.4 - (4 - y) - 2(y + 3) dy = 124.8 J_° 3 (12 + y - y 2 ) dy 
= 124.8 [l2y + t-£] = (-124.8) (-36 + § + 9) = (-124.8) (- f ) = 2808 lb 



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408 Chapter 6 Applications of Definite Integrals 



Using the coordinate system of Exercise 4, we see that the equation for the line of the plate's right-hand edge 
remains the same: y = x — 3 => x = 3 + y and L(y) = 2x = 2(y + 3). The depth of the strip changes to (— y) 

=* F = X.° 3 w(-y)L(y) dy = J°62.4 • (-y) • 2(y + 3) dy = 124.8/j-y 2 - 3y) dy = 124.8 [- y - } 
= (-124.8) (f - |) = (-'W7X2-3) = 561 61b 



If 



-3 



5. Using the coordinate system of Exercise 4, we find the equation for the line of the plate's right-hand edge to be 



2x 



y + 4 



and L(y) = 2x = y + 4. The depth of the strip is (1 — y). 



(a) F = J_° 4 w(l - y)L(y) dy = J^62.4 • (1 - y)(y + 4) dy = 62.4 fjA - 3y - y 2 ) dy = 62.4 Uy - ^ - f 



= (-62.4) [(-4)(4) - 
(b) F = (-64.0) [(-4)(4) - 



(3X16) _ 

2 

(3)(16) 



f) 



+ 



64 



(-62.4) (-16 -24 

_ (-64.0X-120 + 64) _ H94 7 lb 



(-62.4X-120 + 64) 
3 



1164.81b 



6. Using the coordinate system given, we find an equation for 

the line of the plate's right-hand edge to be y = — 2x + 4 

=> x = ^=2 and L(y) = 2x = 4 - y. The depth of the 

strip is (1 — y) =>- F= I w(l— y)(4 — y) dy 



62.4^' (y 2 - 5y + 4) dy = 62.4 [y - ^ + 4y 



(62.4) 



4) = (62.4) 



'2-15 + 24> 




(62.4X11) 



114.41b 



y(ft) 




7. Using the coordinate system given in the accompanying 
figure, we see that the total width is L(y) = 63 and the depth 

of the strip is (33.5 - y) =>• F = / w(33.5 - y)L(y) dy 



J o 33 ^ - (33.5 - y) - 63 dy = (§) (63)/" (33.5 - y) dy 
(§) (63) f33.5y - 



' 64-63 ^ 

v 123 J 



(33.5)(33) - ^ 



(64)(63)(33)(67 - 33) 

(2) (123) 



13091b 



y(in) 

33.51 surface 
"33!ol 




x(in) 



-31.5 



31. S 



(a) Use the coordinate system given in the accompanying 
figure. The depth of the strip is ( ^ — y) ft 

=> F =/" /6 w (if -y) (width) dy 



(62.4)(width)/ 
(62.4)(width) f ' ' 
(62.4)(width) [(f 



/ 6 . 



6 y 2 j 

» F end = (62.4)(2) ( l -§) (1) 

(b) Use the coordinate system given in the accompanying 
figure. Find Y from the condition that the entire volume 
of the water is conserved (no spilling): 4i - 2 - 4 = 2 - 2 - Y 
=> Y = y ft. The depth of a typical strip is ( T - y) ft 
and the total width is L(y) = 2 ft. Thus, 

F = I" 3w (T-y) L oo d y 

11/3 



y(ft) 



i I sur face 



WWW* W 

''y^yy&ffiy&yy ■■:■■:■■:■ 



-l 



Wmm'mmm 



x(ft) 



209.73 lb and F side = (62.4)(4) ($) (1) 



I surface 




x(ft) 



/ "(62.4) (f - y) • 2dy = (62.4)(2) [f y - f ] " = (62.4)(2) [(1 



:^) 2 



419.47 lb 



(62.4)(121) 



838.93 lb 



the fluid 



force doubles. 



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Section 6.7 Fluid Pressures and Forces 409 



Using the coordinate system given in the accompanying 
figure, we see that the right-hand edge is x = y/l — y 2 
so the total width is L(y) = 2x = 2yl — y 2 and the depth 
of the strip is (— y). The force exerted by the water is 

therefore F = I w • (— y) • 2yl — y 2 dy 



62.4/^1 - y 2 d(l-y 2 )= 62.4 [§(1 



,2\3A 




x(ft) 



(62.4) (I) (1 -0) = 416 lb 



10. Using the same coordinate system as in Exercise 15, the right-hand edge is x = ^/3 2 — y 2 and the total width is 
/ 2 . The depth of the strip is (— y). The force exerted by the milk is therefore 



L(y) = 2x = 2^/9 

F = J° 3 w - (-y) • 2 a/9 - y 2 dy = 64.5 /_°y 9 - y 2 d (9 - y 2 ) = 64.5 [§ (9 - y 
= (64.5)(18) = 11611b 



2x3/2 



-3 



(64.5) (?) (27-0) 



11. The coordinate system is given in the text. The right-hand edge is x = */y and the total width is L(y) = 2x = 2,/y. 
(a) The depth of the strip is (2 — y) so the force exerted by the liquid on the gate is F = I w(2 — y)L(y) dy 
= J o '50(2 - y) • 20 dy = 100 £(2 - y)0 dy = lOo/J (2y J / 2 - y 3/ 2 ) d y = 100 [f y 3 / 2 - | y 5 / 2 ] J 



100 I 



(20 - 6) = 93.33 lb 



(b) We need to solve 160 = / w(H — y) • 2^/y dy for h. 



160 



100 (f 



H = 3 ft. 



12. Use the coordinate system given in the accompanying figure. The total width is L(y) = 1. 

(a) The depth of the strip is (3 — 1) — y = (2 — y) ft. The force exerted by the fluid in the window is 

F = £ W (2 - y)L(y) dy = 62.4 £(2 - y) • 1 dy = (62.4) fey - ^1 - <^> ^ (i - ¥\ - <^M 

(b) Suppose that H is the maximum height to which the 
tank can be filled without exceeding its design 
limitation. This means that the depth of a typical 
strip is (H — 1) — y and the force is 

F = j a W [(H - 1) - y]L(y) dy = F nlM , where 



F max = 312 lb. Thus, F max = w£ [(H - 1) - y] • 1 dy = (62.4) |"(H - l)y 




93.6 lb 



x(ft) 



(^) (2H - 3) = -93.6 + 62.4H. Then F max = -93.6 + 62.4H 
6.5 ft 



312 = -93.6 + 62.4H 



H 



405.6 
62.4 



13. Suppose that h is the maximum height. Using the coordinate system given in the text, we find an equation for 



j y. The total width is L(y) = 2x = | y and the 



the line of the end plate's right-hand edge is y 

depth of the typical horizontal strip at level y is (h — y). Then the force is F = I w(h — y)L(y) dy = F max , 

where F max = 6667 lb. Hence, F raax = w£ (h - y) • | y dy = (62.4) (f ) / (hy - y 2 ) dy 

" hy 2 



(62.4) (I) 



Height = h and \ (Base) = I h 



i] o = (62.4) (I) (f - f ) = (62.4) (I) (I) h 3 = (10.4) (*) h 3 => h = \/(f) (ffej) 

f water which the tank can hold is V = | (B 
h 2 ) (30) = 12h 2 « 12(9.288) 2 « 1035 ft 3 . 



!) (Wz) ~ 9 ' 288 ft The volume of water which the tank can hold is V = \ (Base)(Height) • 30, where 



V 



'1u2\ 



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410 Chapter 6 Applications of Definite Integrals 

14. (a) After 9 hours of filling there are V = 1000 • 9 = 9000 cubic feet of water in the pool. The level of the water 



h = jJHj = 6 ft. The depth of the typical horizontal strip at 



is h = ^-, where Area = 50 • 30 = 1500 

Area' 

level y is then (6 — y) for the coordinate system given in the text. An equation for the drain plate's 
right-hand edge is y = x =>• total width is L(y) = 2x = 2y. Thus the force against the drain plate is 

2-1 



(62.4)(2) | \ 



F=J g w(6 - y)L(y) dy = 62.4 J q (6 - y) • 2y dy = (62.4)(2)/ q (6y - y 

= (124.8) (3 - i) = (124.8) (f ) = 332.8 lb 
(b) Suppose that h is the maximum height. Then, the depth of a typical strip is (h — y) and the force 

F = J w(h - y)L(y) dy = F, llilx , where F max = 520 lb. Hence, F max = (62.4) J* (h - y) • 2y dy 



124.8 J o '(hy - y 2 ) dy = (124.8) [ 



hy 2 y J 
~2 T 



(124.8) (| - i) = (20.8)(3h - 2) 



520 
20.8 



3h-2 



h = f = 9 f t 



15. The pressure at level y is p(y) = w • y => the average 
pressure is p = ± j Q p(y) dy = \ J o w • y dy = 1 w j^J 

= (?) (V) = ^t ■ ^ n ' s ' s ^ e P ressure at level | , which 



is the pressure at the middle of the plate. 




16. The force exerted by the fluid is F = I w(depth)(length) dy = I w • y • a dy = (w • a) I y dy = (w • a) y 
= w ( 2 ) = (if) ( a ^) = P ' Area, where p is the average value of the pressure (see Exercise 21). 

/o 

(62.4) (2y 4 — y 2 ) (— y) dy 



(62.4) f° (4 - " 2N i 1/2 ''- ■>-'■ .i> - "'->"> i . I = , i , j , ' - 



y 2 ) 1/z (-2y) dy = (62.4) \i (4 - y 2 ) 



(62.4) (|) (4 3 / 2 ) = 332.8 ft - lb. The force 



3V- 'J j_ 2 -v— ->\m 
compressing the spring is F = lOOx, so when the tank is full we have 332.8 = lOOx => x w 3.33 ft. Therefore 
the movable end does not reach the required 5 ft to allow drainage => the tank will overflow. 



18. (a) Using the given coordinate system we see that the total 
width is L(y) = 3 and the depth of the strip is (3 — y). 

Thus, F = £ w(3 - y)L(y) dy = J fl 3 (62.4)(3 - y) • 3 dy 

= (62.4)(3)/ o 3 (3 - y) dy = (62.4)(3) [3y - £] ' 

= (62.4)(3) (9 - |) = (62.4)(3) (§) = 842.4 lb 




•x(tt) 



-1.5 



1.5 



(b) Find a new water level Y such that F Y = (0.75)(842.4 lb) = 631.8 lb. The new depth of the strip is 
(Y — y) and Y is the new upper limit of integration. Thus, F Y = / w(Y — y)L(y) dy 



62A£(Y - y) • 3 dy = (62.4X3)]^ (Y - y) dy = (62.4)(3) [Yy 
(62.4)(3) (?f\ . Therefore, Y = 



(62.4)(3) ( Y 2 - S 



2F Y 
(62.4X3) 



1263.6 
187.2 



v/6T75 w 2.598 ft. So, AY = 3 - Y 



3 - 2.598 rs 0.402 ft « 4.8 in 



19. Use a coordinate system with y = at the bottom of the carton and with L(y) = 3.75 and the depth of a typical strip being 
(7.75 - y). Then F = /" 5 w(7.75 - V)Hy) dy = ( 6 -§) (3.75)_ j f" 5 (7.75 - y) dy 



1 64.5 > 

V 123 ) 



(3.75) |7.75y- \ 



123 



(3.75) 



4.21b 



Copyright (c) 2006 Pearson Education 




Chapter 6 Practice Exercises 411 



20. The force against the base is F base = pA = whA = w • h • (length)(width) = (^) (10)(5.75)(3.5) « 6.64 lb. 

To find the fluid force against each side, use a coordinate system with y = at the bottom of the can, so that the depth of a 



w(10 — y) 



' width of \ 
. the side / 



dy 



.§)(££) iQy-i 



57 \ /width of N / 100 \ 
12 3 J V the side M 2 / 



2L) (50)(3.5) « 5.773 lb and F side = (S) (50)(5.75) « 9.484 lb 



123; 



21. (a) An equation of the right-hand edge is y 



y and L(y) = 2x 



il 



The depth of the strip 



is (3 - y) =► F = J\(3 - y)L(y) dy = J 3 (62.4)(3 - y) (f y) dy = (62.4) • (£ 



J> 



- y 2 ) dy 



(62.4) (fn |y- 3 



3 „2 



(62.4) (I) [f - f ] = (62.4) (I) (f ) = 374.4 lb 



(b) We want to find a new water level Y such that F Y = \ (374.4) = 187.2 lb. The new depth of the strip is 
(Y — y), and Y is the new upper limit of integration. Thus, Fy = I w(Y — y)L(y) dy 



62.4^ Y (Y - y) (| y) dy = (62.4) (fjJjYy - y 2 ) dy = (62.4) (f) [y 



5l _ y_ 
2 3 



(62.4) (1) (? - f ) 



(62.4) (§) Y 3 . Therefore Y 3 



9Fi 



(9)(187.2) 



3 / (9)(187.2) 



Vl3.5 « 2.3811 ft. So, 



v9 y ^ . i^x.^xw x 2.(62.4) 124.8 ~^ \J 124.8 

AY = 3-Y«3- 2.381 1 « 0.6189 ft « 7.5 in. to the nearest half inch, 
(c) No, it does not matter how long the trough is. The fluid pressure and the resulting force depend only on depth of the 
water. 



22. The area of a strip of the face of height Ay and parallel to the base is 100(||) • Ay, where the factor of || accounts for the 
inclination of the face of the dam. With the origin at the bottom of the dam, the force on the face is then: 

F = /, 24 w(24 - y)(100)(ff)dy = 6760 [24y - £ 1 ** = 676o(24 2 - %f\ = 1,946,8801b. 



CHAPTER 6 PRACTICE EXERCISES 



1. A(x) = 1 (diameter) 2 

= 5( X -2 v /^-x 2 - 



Hv/x-x 2 



0,b= 1 



V = J A(x) dx = I J o (x - 2x 5 /' 2 + x 4 



dx 



|x 7 / 2 + 



4 ^2 



% (35 - 40 + 14) 



4-70 



280 



y=Sx 




2. A(x) = 1 (side) 2 (sin |) 



73~ 



■lyfi. 



TZ 



(4x - 4xa/x + x 2 ) ; a = 0, b = 4 



V = J a A(x) dx = &f o (4x - 4x 3 / 2 + x 2 ) dx 
[2x 2 - f x 5 / 2 4 



73 
4 

32^3 
4 



75 



^(32-^ 



32 | 64 \ 





5^(15-24+10)= ^ 




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412 Chapter 6 Applications of Definite Integrals 



3. A(x) = | (diameter) 2 = | (2 sin x - 2 cos x) 2 
= | • 4 (sin 2 x — 2 sin x cos x + cos 2 x) 
= tt(1 -sin2x);a= f , b = ^ 



A(x)dx = 7r (1 -sin2x)dx 

J-tt/4 

-[(f+^)-(f-^)]=- 2 




4. A(x) = (edge) 2 = N v/6- v/xV-Oj = (v^ - ,/xV = 36 - 24^/6 y'xH- 36x - 4^/6 x 3 / 2 + x : 
a = 0,b = 6 ^ V = £a(\) dx = £ ^36 - 24^ y/x. + 36x - 4^/6 x 3 / 2 + x 2 ) dx 

= [36x - 24v/6 • § x 3 / 2 + 18x 2 - \\fl ■ \ x 5 / 2 + f] = 216 - 16 • v^ v^ - 6 + 18 • 6 2 - § v^ v^ - 6 
= 216 - 576 + 648 - ^ + 72 = 360 - ^ = inoo-ms = n 



2 _,_ §! 



5. A(x) = ^ (diameter) 2 = f (2-v/x - f ) = f (4x - x 5 / 2 + fg ) ; a = 0, b = 4 =4> V = /^(x) 



dx 



[ J o 4 ( 4 x-x5/ 2 + ^)dx=f [2x 2 -fx^ 2 + ^] o =f(32- 32-f + | - 32) 



^(i-f+i; 



|f (35 - 40 + 14) = ^ 



6. A(x) = \ (edge) 2 sin (f ) = f [2^x - {-2y/x)] 



/l 



^ 



4,/x) 2 = 4\/3x; a = 0, b = 1 



>. V = J^AOO dx = f*4y/3 x dx = [2-^3 x 2 ] 

2^ 




y 2 = 4x 




7. (a) rfisfc method: 

V = _£ b 7rR 2 (x) dx = J\ (3x 4 ) 2 dx = tt /V 8 dx 

= TT [X 9 ] _j = 2TT 



(b) sfoeZZ method: 

V = J> (** ) (** ) dx = J>x (3*») dx = 2, - 3 JV dx = 27T - 3 [f ] * = tt 

Note: The lower limit of integration is rather than — 1 . 

(c) shell method: 

v=r 2 -(^)(^) dx = 2 ^ i " x )( 3x4 ) dx = 2 - [¥-Ci =27r til -i) -(-§-1)]=^ 

(d) washer method: 



R(x) = 3, r(x) = 3 - 3x 4 = 3 (1 - x 4 ) => V = f n [R 2 (x) - r 2 (x)] dx = J_ |7 r [9 — 9(1 — x 4 ) 
= 9tt Jjl - (1 - 2x 4 + x 8 )] dx = 9tt J_' 1 (2x 4 - x 8 ) dx = 9n \ 



dx 



2x° ?r 

5 9 



[2 _ 11 _ 2tt-13 _ 267T 
L5 9J 5 5 



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;d 2 



V=/ a ^[R 2 (x)-r 2 (x)]dx = / i ^[(4 s ) : 
'-3)]=^(-i7)-5 + T + ?)=l»(-2-10 + 64 + 5) 



dx = 2tt -4x _1 



M(- 



(a) washer method: 
R(x) = 4j , r(x) = \ 

= *[($-*)-■ 

(b) s/ieZZ method: 

(c) s/ieZZ method: 

V = ar^CSl) (** ) dx = 2 -r^ 2 " x ) (P - 5) dx = 2»jT(i 
= 2tt [- I + f -x + f ] 2 = 2tt [(-1 + 2 - 2+ 1) - (-4 + 4 - 1 

(d) washer method: ? 

V = / tt[R 2 (x) - r 2 (x)] dx 



Chapter 6 Practice Exercises 

dx = 7T [- f X- 5 



413 



x] 2 
4 J 1 



57- 
20 



I)} 



2- (I) 



5jr 
2 



-£-1 



dx 



11 - 16ttJ] (1 -2x 



-3 , v -6\ 



dx 



49?r 
4 

49?r 

4 
49jr 

4 
49jr 

4 



167T X + X" 



-2 X" 



16tt[(2- 



16tt 

[far 

160 



'1 
(40 



1 _ 

" 4 

1 

160 



5-32^ 
4) 



-(1 + 1-1)] 



1 + 32) 



49.-; 
4 



7_hr 
10 



103_7T 

20 




2- x 



(a) rfisfc method: 

V = 7T J] (a/x- l) dx = 7T J] (X - 1) dX = 7T [f - xl 

= *[(¥-5)-(£-l)]=*(¥-4)=8* 

(b) washer method: 

R(y) = 5, r(y) = y 2 + 1 => V = £ tt [R 2 (y) - r 2 (y)] dy = tt £ [25 - (y 2 + l) 2 ] dy 

= njj25 - y 4 - 2y 2 - 1) dy = tt £(24 - y 4 - 2y 2 ) dy = tt [24y - £ - f y 3 ] ^ = 2tt (24 - 2 - f - f 

= 32tt (3 - § - i) = ^ (45 - 6 - 5) = ^ 

(c) rfisfc method: 
R(Y) = 5 - (y 2 + 1) = 4 - y 2 

=> V = J ttR 2 (y) dy = f_n (4 - y 2 ) 2 dy 

= tt J^(16-8y 2 + y 4 )dy 

= 7r[l6y-f + fl 2 =27r(32-f + f ) 




64tt (1 - § + 5) = ^ (15 -10 + 3) 



5127T 

15 



10. (a) shell method: 

v = £ 2, (** ) (*J) dy = />y (y - J) dy 

= 2 -r(y 2 ^) d y= 2 -[T-^;= 2 -(f-f) 



2- 



_ . 64 = ^ 

12 UH 3 




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414 Chapter 6 Applications of Definite Integrals 



(b) shell method: 

V = J> ( ** ) ( **) dx = />x ( 2v /x - x) dx = 2. £(2x 3/2 ~ x 2 ) dx = 2n [| x^ - f ] * 

= 27r(|-32-f) = if£ 

(c) s/ieZZ method: 

v = j> (it ) (it) dx = r^ 4 - *) ( 2 v^ - x ) dx = 2 -x> xi/2 - 4x - 2 * 3/2 + x2 ) dx 

= 2ir [f x 3 / 2 - 2x 2 - § x 5 / 2 + f ] 4 = 2^ (4 



, ,-32-f-32+f)=647r(f-l-f + §) 



64ir (1 - |) = %?■ 



(d) sfoeZZ method: 



V = r2.(^»j( h t;: 1 i ; t )dy = />(4-y)(y-^)dy = 2.X 4 (4y-y 2 -y 2 + ^)dy 
= 27r/ o 4 (4y - 2y 2 + {) dy = 2^ [2y 2 - f y 3 + g] * = 2tt (32 - § - 64 + 16) = 32tt (2- f + l) 



32- 
3 



11. disk method: 

tan J x dx = 7r I (sec x — 1) dx = 7r[tan x — x] 



tt/3 _ t(3V3-7tj 



12. disfc method: 

V = 7rJ (2 - sin x) 2 dx = 7T J (4-4 sin x + sin 2 x) dx = ttJ (4-4 sin x + '~ c ° s2x ) dx 
= tt [4x + 4 cos x + | - 22j2s] * = 7T [(4tt - 4 + f - 0) - (0 + 4 + - 0)] = tt (f - 8) = f (9tt - 16) 



13. (a) disk method: 



32 



V = tt/ o (x 2 - 2x) 2 dx = 7T J o (x 4 - 4x 3 + 4x 2 ) dx = tt [^ - x 4 + f x 3 j = tt v , 



16+ f) 



^ (6 - 15 + 10) = ^ 



2 _ 8tt 
5 — 5 



(b) washer method: 

V = / Vl 2 - (x 2 - 2x + l) 2 j dx = /Vdx + f\ (x - l) 4 dx = 2tt - [tt &=£] = 2tt - 

(c) shell method: 

V = /> (■» (fiSt) dx = 2^(2 - x) [- (x 2 - 2x)] dx = 2^(2 - x) (2x - x 2 ) dx 

= 2tt£(4x - 2x 2 - 2x 2 + x 3 ) dx = 2tt J^(x 3 - 4x 2 + 4x) dx = 2n H - f x 3 + 2x 2 j = 2tt (4 - f + 8 

= f (36 - 32) = f 

(d) washer method: 

V = tt Jj2 - (x 2 - 2x)] 2 dx - tt£ 2 2 dx = tt £ U-4 (x 2 - 2x) + (x 2 - 2x) 2 j dx - 8tt 
= nf (4 - 4x 2 + 8x + x 4 - 4x 3 + 4x 2 ) dx - 8tt = tt J q (x 4 - 4x 3 + 8x + 4) dx - 8tt 



tt | ^ - x 4 + 4x 2 + 4xT - 8tt = tt (f - 16 + 16 + 8) - 8tt = f (32 + 40) - 8tt = ^ - ^ = ^ 



14. disfc method: 

"ir/4 



V = 2?r JJ 4 tan 2 x dx = 8tt J* (sec 2 x - 1) dx = 8?r[tan x - x]q /4 = 2tt(4 - tt) 



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Chapter 6 Practice Exercises 415 



15. The material removed from the sphere consists of a cylinder 
and two "caps." From the diagram, the height of the cylinder 



is 2h, where h 2 



2 2 ,i.e. h= l.Thus 



V cy i = (2h)7r( y3 j = 6tt ft 3 . To get the volume of a cap, 
use the disk method and x 2 + y 2 = 2 2 : V cap = I 7rx 2 dy 
= / i 2 7r(4-y 2 )dy = ^[4y-^] i 2 
= tt[(8 - f) - (4- i)] = r f ft 3 . Therefore, 



* removed — * cyl T" ^ V ca p — D7T 



12ZL — 28tt f t 3 
3 — 3 ii . 




16. We rotate the region enclosed by the curve y = J 12 (1 



4x2n 

121/ 



and the x-axis around the x-axis. To find the 



volume we use the disk method: V = J ^ 7rR 2 (x) dx = J t: (v/12 (l - y§|) J dx = 7r J \2 M - yfM dx 

= 12 -lt( 1 -^) dX=12 "[ X 



4x J 
363 



247T |^ 



' 4 \ / JJ_\ 3 
V363/ 1 Ui 



'32., !-(&)(£) 



1327T 1 



264t 
3 



3tt w 276 in 



3 



17- y = xV 2 -^^ | = i x -i/2_I xl / 2 ^ (|) =|(^-2 + x) =.L=/ i 7l + H^-2 + x)dx 

=> L = J i 4 v /l(I + 2 + x)dx = J^ (X-1/2 + x l/2) 2 dx = J" I ( x -l/2 + x l/2) dx = I [ 2x l/2 + 2 x 3/2] J 

= |[(4+l-8)-(2+|)]=i(2+f) 



3 



18. x = y 2 / 3 => ^ = lx- 1 /3 



dy 



(!)'■ 



4x~ 2 / 3 
9 



L =XV 1+ (I) * = I V 1 + ^ * 



r 



\/9xV3 + 4 
3xV3 



dx = 5/V9X 2 / 3 + 4 (x- 1 / 3 ) dx; [u = 9x 2 / 3 +4 => du = ey" 1 / 3 dy; x = 1 => u = 13, 



X = 8 => U = 40] -+ L = i £° U V2 du = _L [| u 3/2] JO = J. [ 40 3/2 _ ^3/2] M 7^34 



19. y = 5. x 6/5 _ 5 x 4/5 => g = 1 x 1 / 5 - 1 X-VB =* (g)" = I (x 2 / 5 - 2 + X" 2 / 5 ) 



=> L=/ Jl + i(x 2 / 5 -2 + x- 2 / 5 )dx => L = / Ji (x 2 / 5 + 2 + x- 2 / 5 ) dx = / v/± ( X V5 + x -!/5) 2 dx 



j; 32 I(x 1 / 5 +x-V 5 )dx=I[^ 6/5 + ix^]f=I[(f.2« + |.2^)-(f + 5)] = I(^ + -) 
1(1260 + 450)=^ = ^ 



48 



20. x=iy 3 + i => | = |y 2 -^ 



(I) =i^y 4 ^ + ^L=/Vl + ( 1 Ly^I + ^)dy 



f;^ 6 y 4 + ^^y = fJ{ly 2 + ^~«y = L{h 2 + ^«y={hy 3 -l] 1 



(3.- i) _ (1.-1) = 2. + i - 13 

V 12 2^ v 12 ^ 12 T 5 



12 



21. f t = -5 sin t + 5 sin 5t and f 



5 cos t — 5 cos 5t 



, dt) 



(*)' 



(—5 sin t + 5 sin 5t) + (5 cos t — 5 cos 5t) 



5 v sin 2 5t — 2sin t sin 5t + sin 2 1 + cos 2 1 — 2cos t cos 5t + cos 2 5t = 5 \Jl — 2(sin t sin 5t + cos t cos 5 t) 
5^/2(1 - cos it) = 5 J 4(\) (1 -cos4t) = lO^sin 2 2t = 10|sin 2t| = lOsin 2t (since < t < | ) 



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416 Chapter 6 Applications of Definite Integrals 



Length = J^ 10sin2tdt = [-5cos2t]o /2 = (-5)(-l) - (-5)(1) = 10 



22. | =3t 2 - 12tand^ 



3t 2 + 12t^ J(f) + (if J = y / (3t 2 -12t) 2 + (3t 2 + 12t) 2 = ^28812 + 18t 4 
3 v / 2|tl\A6+l 2 ^ Length = J Q 3^ |t|-/l6 + t 2 dt = 3^^ t v / 16+l 2 dt; [u = 16 + t 2 =>• du = 2t dt 



=* idu = t dt; t = => u = 16; t = 1 => u = l?j ; ^ / ^du = ^ [|u 3/2 ] JJ = ^ (§(17) 3/2 - f (16) 



3/2 



h£ . §((17) 3/2 - 64) = ^l({ll) m - 64) 



8.617. 



23. % = -3 sin and & 



:! cos w = 4 / ( £ ) - + ($j\ = y (-3 sin (9) 2 + (3 cos 6») 2 = v/3(sin 2 61 

3ir/2 r-3ir/2 



cos 2 (?) 



/>3ir/2 /.37T/2 

Length =J o 3d0 = 3j o &6 = 3(f - 0) 



2 



24. x = t 2 and y = f - t, -a/3 < t < \/3 =4> 



2t and 



dy 



•n/3 



■n/3 



t 2 - 1 => Length 

•V3 



J-V3 

•V5 



(2t) / + (t 2 -l) z dt 



4x/3 



^3 



-V3 



25. Intersection points: 3 — x 2 = 2x 2 =>• 3x 2 — 3 = 

=>- 3(x - l)(x + 1) = =4> x = -1 orx = 1. Symmetry 
suggests that x = 0. The typical vertical strip has 

center of mass: (x $ ) = ( x, + ( 2 ~ x - J = ( x, 2L -±^ J , 

length: (3 - x 2 ) - 2x 2 = 3 (1 - x 2 ), width: dx, 
area: dA = 3(1— x 2 ) dx, and mass: dm = 6 ■ dA 
= 36 (1 — x 2 ) dx =4> the moment about the x-axis is 

y dm= |<S(x 2 + 3)(l-x 2 )dx= \8 (-x 4 - 2x 2 + 3) dx => M x = / y dm = \ 6 J_ t (-x 4 - 2x 2 + 3) dx 




l4-f-¥+3x 



35 (- i - I + 3) = fS (-3 - 10 + 45) 



35 x 



5 3 

66(1 - i) =45 ^ y 



3^ 
15 



32.^ 
5 



,M= Jdm= 35/^(1 -x 2 )dx 



Ml = |M _ £ _ Therefore, the centroid is (x,y) = (0, f ) 



26. Symmetry suggests that x = 0. The typical vertical 
strip has center of mass: ( x , y ) = (x, f ) , length: x 2 , 



width: dx, area: dA = x 2 dx, mass: dm = S - dA = <5x 2 dx 

s 

2 



the moment about the x-axis is y dm = | x 2 • x 2 dx 



x 4 dx =>• M x = / y dm = § f x 



: dx=^[x^ 



26 (o5\ 
10 V 4 i 



32A 



M 



/dm = 5 J_; 2 x 2 dx = 6[f] 




f(2 3 ) 



y = m = tm = s ■ Therefore, the 



centroid is(x, y) = (0, 



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Chapter 6 Practice Exercises 417 



27. The typical vertical strip has: center of mass: (x , y ) 

4+^\ , ... .2 



y=(l/4)x2 



x, -^f- j , length: 4 — \, width: dx, 

area: dA = (4 — ^- )dx, mass: dm = 5 • dA 
= 6 (4 — ^ ] dx => the moment about the x-axis is 

y dm = 6 • ^^ (4 - f ) dx = § (l6 - fg) dx; the 

moment about the y-axis is 3c dm = <5 ( 4 — ^ I • x dx = <5 ( 4x — ^ J dx. Thus, M x = J y 1 dm = § I ( 16 — h ) dx 

= f[ 1 6x-5 5 1 6]^ = f[64-f]=^;M y = /xdm = ^X 4 (4x-f)dx=^[2x2-fl]^ 

= 5(32- 16)= 166; M = J dm = 6 £U- f) dx = 6 Ux- f^l = 6 (16 



64\ 
12 J 



325 
3 



M " 


16-6-3 
" 32-5 ~ 


= j and y = 


M x 

M 


128-5-3 

" 5-32-5 



12 



Therefore, the centroid is (x, y) 



'3 m 

v2' 5 ) 



28. A typical horizontal strip has: 

center of mass: (3c , y 1 ) = ( y %~ y , y ) , length: 2y — y 2 , 

width: dy, area: dA = (2y — y 2 ) dy, mass: dm = 6 ■ dA 
= S (2y — y 2 ) dy; the moment about the x-axis is 

y dm = 6 - y • (2y — y 2 ) dy = S (2y 2 — y 3 ) ; the moment 

about the y-axis is 3c dm = 6 - ^ y | y ^ • (2y — y 2 ) dy 
= f (4y 2 - y 4 ) dy => M x = / y dm = 6 Jj2y 2 - y 3 ) dy 

2 „3 y 4 l _ i. (2 o 16 



5 (±8 _ 32-1 
2 V 3 5 ) 








nt-s-f 



*(¥-¥) 



5-16 
12 



-•M 



325 



M 



J"3cdm=fJ>y 2 -y 4 )dy = f [ 

Jdm = 4 2 (2y-y 2 )dy = 5 [y 2 - ^ = £ (4 - § 



45 
3 



M. 
M 



4-5-3 
3-4-5 



1. Therefore, the centroid is (x, y) = (|, l) . 



My 

M 



4 y 3_ 

5-32-3 
15-5-4 



and 



29. A typical horizontal strip has: center of mass: (3c , y ) 

= f^ir^y) . len g th: 2 y - y 2 > width: dy, 

area: dA = (2y — y 2 ) dy, mass: dm = 8 ■ dA 

= (1 + y) (2y — y 2 ) dy =>- the moment about the 
x-axis isy dm = y(l + y) (2y — y 2 ) dy 

= (2y 2 + 2y 3 - y 3 - y 4 ) dy 

= (2y 2 + y 3 — y 4 ) dy; the moment about the y-axis is 

3c dm = (^) (1 + y) (2y - y 2 ) dy = \ (Ay 2 - y 4 ) (1 + y) dy = \ (4y 2 + 4y 3 




y 4 - y 5 ) dy 



M 



/ y dm = J^(2y 2 + y 3 - y 4 ) dy = [f 



2 y 3 



]6 i ]6 

J "t" A 



w, 



i6 a + 



|§ (20 +15 -24) =£(11) 



44. M 

15 ' lvl y 



l( 4 #+2 4 

T(2y + y 2 - 

(S)(i)H 



5 



f)=4(| 



4 1 9 _ 4 _ 8\ 
z 5 b) 



/3c dm = £\ (4y 2 + 4y 3 - y 4 - y 5 ) dy = \ [f y 3 + y 4 - £ - 
f ; M = /dm = / o 2 (1 + y) (2y - y 2 ) dy 



y 3 )dy 



(4- 



4(2- J) 



T> 



*=% = &) 



fandy=^ 



|4 . Therefore, the center of mass is (x, y) 



'9 n \ 
v5' 10J 



30. A typical vertical strip has: center of mass: (3c , y ) 



area: dA 



3 



dx, mass: dm = 8 - dA 



^■2 



dx 



*' 2xV2. 

the moment about the x-axis is 



, length: 



-Ik , width: dx, 

X J /2 ' 



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418 Chapter 6 Applications of Definite Integrals 



'y dm = ^4/2 ' b-m dx = Pj dx; the moment about the y-axis is x" dm = x • tf-fU dx = SrL dx. 



2x 3/2 W x 3/2 
•>9 



x'- 1 



(a) M x = 5 j;i(|)dx=f [-^] i = ^;M y = 4 i x(^)dx = 35[2xV^ = 126; 
M = ^ 9 ^dx=^[x-V 2 ]; =4 ^x=t = ^ = 3andy=^=(l)=f 

(b) M x = jj (|) dx = § [- I] J = 4; M y = J> (£) dx = [2^] J = 52; M = £ x , 

= 6[xV^ = 12^ J .= ^ = fandy=t = | 



x" - 



dx 



3i. s = £2«ysJi + (%) dx;| = ^TT =* (I) =2^1 ^S = X 3 2 7 r^/2xTT^l + 2l ^ T dx 

= 27r/ 3 v^ + T yM dx = 2^/ o 3 v^TT dx = 2^ [§ (x + l) 3 / 2 ] I = ly/li: . § (8 - 1) = ^2 



3ZS = / i 2«y^l+(g) d*;|=x»=>(g) = x* =► S = J* o 2tt - f v^T^dx = |/ o v'TT^ (4x3) 
= I X l ^/^T^d(l +x^) = 1 [f (1 + x 4 ) 3/2 ] ^ = f [2V2- 1] 



dx 



. S = />x^/l + (|) dy 



dx _ (|)(4-2y) _ 2-y 



dy ' ^4y^2 " ,/4y~^ 
>2 



(I) 1 



4y - y 2 + 4 - 4y + y 2 

4y_y2 



4y-r 



dy = 47r I dx = 47r 



4y-y 2 



34. S = />x^l + (I) dy; | = ^ =* 1 + (|) = 

= ^/ 2 6 y47TTdy = f [f (4y + l) 3 / 2 ] J = f (125 - 27) = | (98) 



J_ _ 4y+l 
4y — 4y 



T 2 Vy- 



v / 4y+T 
x/4y 



dy 



-IS).r 

3 



35. x = f and y = 2t, < t < y/l => f t = t and ^ = 2 => Surface Area = J o 5 27r(2t) v / t 2 + 4 dt = J]' 27TU 1 / 2 du 



2. [§ U 3/ 2 ] ; 



76»r 



where u = t 2 + 4 => du = 2t dt; t = => u = 4, t = x/5 => u 



36. x = t 2 + I and y = 4^, 4- < t < 1 =* f = 2t - ^ and $ 



Surface Area = J ' 2ir (t 2 + i 



2t-^) 2 + 



($) * = 2 */. 



M v 



;2t+^dt 



= 2 - O t2 + r t ) (» + & dt = 2, j;^ +1+1 1 -3) dt = 2. [i t* + 1 1 - 1 1 - 2 ] ; /V5 
= 2.(2-^) 

37. The equipment alone: the force required to lift the equipment is equal to its weight => Fi(x) = 100 N. 

Xb />40 

Fi(x) dx = I 100 dx = [100x]q° = 4000 J; the rope alone: the force required 

to lift the rope is equal to the weight of the rope paid out at elevation x =>• F2(x) = 0.8(40 — x). The work 



Xb /*40 r 

F 2 (x)dx=J () 0.8(40 -x)dx = 0.8 40x 

the total work is W = W 1 + W 2 = 4000 + 640 = 4640 J 



40 



0.8 



(40 2 - f ) 



(0.8X1600) 



640 J; 



38. The force required to lift the water is equal to the water's weight, which varies steadily from 8 • 800 lb to 

8 • 400 lb over the 4750 ft elevation. When the truck is x ft off the base of Mt. Washington, the water weight is 



F(x) = 8 - 800 - ^ 



2-4750 - x x 



2-4750 



(6400) (1 



9500 



lb. 



The work done is W = I F(x) dx 



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L 

Jo 

22,800,000 ft • lb 



6400 (1- g&j) dx 



6400 x- 



2-9500 J 



47o() 



6400 4750 



4750 2 
4-4750 



Chapter 6 Practice Exercises 

) = (|) (6400)(4750) 



419 



39. Force constant: F = kx =>■ 20 = 
W = J o kx dx = k f o x dx = |20 



k- 1 

1 



> k = 20 lb/ft; the work to stretch the spring 1 ft is 
10 ft - lb; the work to stretch the spring an additional foot is 



W = f* kx dx = k f\ dx = 20 [f ] \ = 20 ( I |) 



20 i 



30 ft • lb 



40. Force constant: F = kx =>■ 200 = k(0.8) =4> k = 250 N/m; the 300 N force stretches the spring x = | 

= |^g = 1.2 m; the work required to stretch the spring that far is then W = I F(x) dx = I 250x dx 
= [125x 2 ]i- 2 = 125(1.2) 2 = 180 J 



41. We imagine the water divided into thin slabs by planes 
perpendicular to the y-axis at the points of a partition of the 
interval [0, 8]. The typical slab between the planes at y and 
y + Ay has a volume of about AV = 7r(radius) 2 (thickness) 



=& y 2 Ay ft 3 . The force F(y) required to 



= t(!y) Ay 
lift this slab is equal to its weight: F(y) = 62.4 AV 
_ (62.4X25) 7ry2 Ay lb The distance through which F(y) 

must act to lift this slab to the level 6 ft above the top is 

about (6 + 8 — y) ft, so the work done lifting the slab is about AW 



y 

14 


8 






/^ x 4 " 







-10 10 

Reservoir's Cross Section 



(62.4)(25) 
16 



7ry 2 (14 - y) Ay ft • lb. The work done 



lifting all the slabs from y = to y = 8 to the level 6 ft above the top is approximately 



W«E 



(62.4)(25) 



[(S 7ry (14 — y) Ay ft • lb so the work to pump the water is the limit of these Riemann sums as the norm of 



the partition goes to zero: W = I ( ' )( - 



/o (16) Vd4-y)dy 
(62.4) (^f) (t • g3 ~ f ) ~ 418,208.81 ft • lb 



(62^25), ^(^2 



y 3 )dy = (62.4)(^)[fy 3 -£] 



42. The same as in Exercise 41, but change the distance through which F(y) must act to (8 — y) rather than 
(6 + 8 — y). Also change the upper limit of integration from 8 to 5. The integral is: 

w = | o 5 «^p y2(8 _ y) dy = (62 4) (2^) ^ (8y2 _ y3) dy = (624) (25.) J 8 y 3 _ t] * 

= (62.4)(^)(f.5 3 -?) 



54,241.56 ft -lb 



43. The tank's cross section looks like the figure in Exercise 41 with right edge given by x = 4, y = ?. A typical 



horizontal slab has volume AV = 7r(radius) 2 (thickness) 



'Ay 



I y 2 Ay. The force required to lift this 



slab is its weight: F(y) = 60 • | y 2 Ay. The distance through which F(y) must act is (2 + 10 — y) ft, so the 
work to pump the liquid is W = 60 J o tt(12 - y) (^) dy = 15tt [^ - £1 = 22,500tt ft • lb; the time needed 



to empty the tank is f^°°™ 



257 sec 



44. A typical horizontal slab has volume about AV = (20)(2x)Ay = (20) (2a/16 — y 2 ) Ay and the force required to 
lift this slab is its weight F(y) = (57)(20) (2-^/16 — y 2 ) Ay. The distance through which F(y) must act is 
(6 + 4 — y) ft, so the work to pump the olive oil from the half-full tank is 

W = 57/_° 4 (10 - y)(20) (2^16 - y 2 ) dy = 2880 f_ 4 10^16 - y 2 dy + 114o/_° 4 (16 - y 2 ) 1/2 (-2y) dy 



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420 Chapter 6 Applications of Definite Integrals 

= 22,800 • (area of a quarter circle having radius 4) + \ (1 140) [(16 - y 2 ) 3/2 j = (22,800)(4tt) + 48,640 



335,153.25 ft -lb 



45. F = £w • ( d s ^ t p h ) • L(y) dy => F = 2 £(62.4X2 - y)(2y) dy = 249.6£(2y - y 2 ) dy = 249.6 [y 



2 _ r 



(249.6) (4 - f) = (249.6) (f) = 332.8 lb 



46. F = JV - (**) - L(y) dy => F = £ 6 ?5 (| - y) (2y + 4) dy = 75 £ '(§ y + f - 2y 2 - 4y) dy 



75 X 5/ °(f-ly- 2 y 2 )dy = 75[fy-^ 2 - 2 y3]f = (75)[(f§)-(|)(|)-(f)(l)] 

(75)(3075) 



(75) 



25 _ 175 _ 250 

9 216 3-216^ V9-2 



^g) (25 -216 -175-9-250-3)- „..,,„ 



k6^ \36J \3J \216J 

118.631b. 



47. F = JV - ( d ^ t p h ) • L(y) dy =* F = 62.4^ (9 - y) (2 - ^) dy = 62a£ (9y 1 / 2 - 3y 3 / 2 ) dy 

= 62.4 [6y 3 / 2 - I f' 2 } J = (62.4) (6 • 8 - § • 32) = (^) (48 - 5 - 64) = ^^2 = 2196.48 lb 

48. Place the origin at the bottom of the tank. Then F = I W • ( d ^ n t p h J • L(y) dy, h = the height of the mercury column, 



strip depth = h - y, L(y) = 1 =► F = £849(11 - y) 1 dy = (849) f g (h - y) dy = 849 [hy - f 1 = 849 (h 2 - f ) 
= ^h 2 . Now solve ^h 2 = 40000 to get h w 9.707 ft. The volume of the mercury is s 2 h = l 2 • 9.707 = 9.707 ft 3 . 

F = wi/ o 6 (8 - y)(2)(6 - y) dy + w 2 £(8 - y)(2)(y + 6) dy = 2wi£(48 - 14y + y 2 ) dy + 2w 2 £.(48 + 2y - y 2 ) dy 



2wi |48y - 7y 2 + \ \ ] + 2w 2 |48y + y 2 - £ | = 216wi + 360w 2 



50. (a) F = 62.4/ o 6 (10-y)[(8 



I) - (I)] dy 



62.4 



/ (240 - 34y + y 2 ) dy 



y »6x 



^ 240y-17y 2 + ^ =^(1440-612 + 72) 



18,7201b. 



(]fi) (7.6) 

y » -€x + 48 



(7/2,3) • 



(6,0) 



(8.0) 



(7.2) 



(b) The centroid ( 5 , 3) of the parallelogram is located at the intersection of y = = x and y = — | x -+- 4r . The centroid of 



6 v 1 36 
5 X -I- 5 . 



the triangle is located at (7, 2). Therefore, F = (62.4)(7)(36) + (62.4)(8)(6) = (300)(62.4) = 18,720 lb 
CHAPTER 6 ADDITIONAL AND ADVANCED EXERCISES 

1. V = 7T Jjf(x)] 2 dx = b 2 - ab => 7r£[f(t)] 2 dt = x 2 - ax for all x > a =^> tt [f(x)] 2 = 2x - a =>• f(x) = ± J^f- 

2. V = 7T f [f(x)] 2 dx = a 2 + a =>■ tt f* [f(t)] 2 dt = x 2 + x for all x > a => ?r[f(x)] 2 = 2x + 1 =>- f(x) = ± 

•J *J 



2x+ 1 



3. s(x) = Cx => £ v/l + [f'(t)] 2 dt = Cx => a/1 + [f'(x)] 2 = C => f'(x) = ^C 2 - 1 for C > 1 

=$► f(x) = f* VC 2 - 1 dt + k. Then f(0) = a => a = + k => f(x) = f ^C 2 - 1 dt + a => f(x) = x^C 2 - 1 + a, 
where C > 1 . 



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Chapter 6 Additional and Advanced Exercises 421 

4. (a) The graph of f(x) = sin x traces out a path from (0, 0) to (a, sin a) whose length is L = / \J 1 + cos 2 8 d(9. 

The line segment from (0, 0) to (a, sin a) has length y/(a — 0) 2 + (sin a — 0) 2 = \J o? + sin 2 a. Since the 
shortest distance between two points is the length of the straight line segment joining them, we have 

immediately that J y/l + cos 2 9 dd > \J a 2 + sin 2 a if < a < § . 

(b) In general, if y = f(x) is continuously differentiable and f(0) = 0, then I y/l + [f'(t)] 2 dt > \Ja 2 + f 2 (a) 
for a > 0. 



5. From the symmetry of y = 1 — x", n even, about the y-axis for — 1 < x < 1 , we have x = 0. To find y 



M 



we 



use the vertical strips technique. The typical strip has center of mass: (x , y ) 



l-x'M 



, length: 1 — x", 



width: dx, area: dA = (1 — x n ) dx, mass: dm = 1 • dA = (1 — x") dx. The moment of the strip about the 



x-axis is y dm 



„n>2 



1 



(n+l)(2n+l)-2(2n + l) + (n+l) 



€i0 



dx = 2| i(T - 2x n + x 2n ) dx = |x- 

2n 2 + 3n + I - 4n - 2 + n + 1 _ 2n 2 



n+1 



x-° + ' 1 1 

2n + I J 



n+1 



2n+l 



(n+l)(2n+l) 



(n+l)(2n+l) 



(n+l)(2n+l) 



Also, M = /_'dA = Jjl - x») dx = 2 X'(l - x») dx = 2 [x - ^] J = 2 (1 



n+ 1/ 



2n 



M 



2n 2 



(n+1) 



(n+l)(2n+l) 2n 2n+l "^ 

the limiting position of the centroid is (0, \ 



o. 



2n+l 



is the location of the centroid. As n 



n+1 

* oo, y 



Therefore 

l 



so 



6. Align the telephone pole along the x-axis as shown in the 
accompanying figure. The slope of the top length of pole is 



/ N.5 _ 9 \ 
V St i-) 



40 



8tt 40 



(14.5 - 9) 



5.5 
8?r-40 



8ri-S(> 



. Thus, 



H) 



8- 



8,7 



y = SF + 8^0 x = t ( 9 + So x ) is an equation of the 



(*■*) 



y-^ 9 *^ 



80 



\ 




(«3r) 



■*■ x 



(«-W) 



line representing the top of the pole. Then, 

x-^y 2 dx = 7rJ o x[i(9+|ix)] dx 
= 64-],, ^ (9 + ^ x) dx;M=J ^y 2 dx 

= " X, 40 [ si ( 9 + 35 x ) ] 2 dx = 5S r" ( 9 + SB x ) 2 dx - Thus ' x = t « fif ~ 23 06 (usin S a caIculator t0 com P ute 
the integrals). By symmetry about the x-axis, y = so the center of mass is about 23 ft from the top of the pole. 



(a) Consider a single vertical strip with center of mass ( x , y ). If the plate lies to the right of the line, then 
the moment of this strip about the line x = b is (x — b) dm = (x — b) 6 dA => the plate's first moment 
about x = b is the integral J (x - b)<5 dA = J <5x dA - J* <5b dA = M y - b<5A. 

(b) If the plate lies to the left of the line, the moment of a vertical strip about the line x = b is 

(b — 3c ) dm = (b — 3c ) S dA => the plate's first moment about x = b is J (b — x)<5 dA = J bS dA — J <5x dA 
= b<5A - M„. 



(a) By symmetry of the plate about the x-axis, y = 0. A typical vertical strip has center of mass: 

(x ,y ) = (x, 0), length: 4^/ax, width: dx, area: 4^/ax dx, mass: dm = S dA = kx - A^fax dx, for some 

proportionality constant k. The moment of the strip about the y-axis is M y = J 3c dm = / 4kx 2 ^/ax dx 

= 4k,/aJ" X 5 / 2 Hx = 4k, A n X 7 / 2 ! a = 4 ka l/2 . 2 a 7/2 = 8kal 



' x 5 / 2 dx = 4k ^a [f x 7 / 2 ] * = 4ka 2 / 2 • i a 7 / 2 = ^ . Also, M = J dm = J" 4kx ^ax dx 



= 4k v /^/ o a x 3 / 2 dx = 4k^ [§ x 5 / 2 ] J = 4ka J / 2 . § a 5 / 2 = ^ . Thus, x 
=> (x, y) = (y , 0) is the center of mass. 

(b) A typical horizontal strip has center of mass: (x ,y ) = I 4a 2 a , y J = ( 



8ka 4 



Ska ' 



y 2 +4a 2 
8a ' 



y ) , length: a - fj 



width: dy, area: ( a — J- J dy, mass: dm = 6 dA = |y | (a — |- J dy. Thus, M x = J y dm 

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422 Chapter 6 Applications of Definite Integrals 



a „3 r 

20a 



M 



£ y ly I (a - i) dy = £ a -y 2 (a - g) dy + £ y 2 (a - g) dy 

/l(-ay 2 + ^)dy + /;(ay 2 -^)dy=[ 

-¥ + ^ + ¥"S = 0:M, = /*dm=£;(^!)|y|(a-g)dy 

i / Jy| (y 2 + 4a 2 ) (*^) dy = ^ / |y| (16a 4 - y 4 ) dy 

V ' -2a 

3^£ a (-16a 4 y + y 5 )dy+3^X 2a (16a 4 

jfc [8a 4 - 4a 2 - if] + £ [ 8 a 4 - 4a 2 - if] = £ ( 3 2a' 

= / dm = iIiy|(^) d y = siIiy|( 4a2 -y 2 ) d y 
s l_J- 4a2 y + y 3 ) d y + k C( 4a2 y - y 3 ) d y = k [- 2a 

2 • i ^2a 2 • 4a 2 - ±f-*) = i (8a 4 - 4a 4 ) = 2a 3 . Therefore, x 



1 y-y 5 )dy=3^|-8a 4 y 2 ~>- 



+ ^P 8-V-J 



,6 _ 32a 6 



,2 y 2 + £ 



-2a 

sP ' 3 (32a ) = 3 a ; 



!2a 2 y 2 -^ 



M 



)(£) 



°f and 



y = ^ = is the center of mass. 



9. (a) On [0, a] a typical vertical strip has center of mass: (x , 'y ) = ( x, — — ~ x ^ a ~ x j , 

length: y b 2 — x 2 — ya 2 — x 2 , width: dx, area: dA = I y b 2 — x 2 — y a 2 — x 2 J dx, mass: dm = 6 dA 
= 6 I v b 2 — x 2 — v a 2 — x 2 J dx. On [a, b] a typical vertical strip has center of mass: 
(x , y ) = (x, ^ 



mass: 



length: y b 2 — x 2 , width: dx, area: dA = y b 2 — x 2 dx, 
dm = 6 dA = S ^b 2 - x 2 dx. Thus, M x = / y dm 
£\ ( v / b 2 -x 2 + Va 2 - x 2 ) 6 (Vb 2 -x 2 - a/ a 2 - x 2 ) dx + _£" 1 x/b 2 - x 2 6 ^b 2 - x 2 dx 



Jj(b 2 - x 2 ) - (a 2 - x 2 )] dx + | J* (b 2 - x 2 ) dx = | Jjb 2 - a 2 ) dx + § J* (b 2 - x 2 ) dx 
[(b 2 - a 2 ) x] a + f [b 2 x - f ] " = | [(b 2 - a 2 ) a] + § [(b 3 - f ) - (b 2 a - f )] 
(ab 2 - a 3 ) + f (§ b 3 - ab 2 + f ) = ^ - f = «5 (^) ; M y = / x dm 
J\o (yjb 2 -x 2 - A/a 2 - x 2 ] dx + f\d s/b 2 - x 2 dx 



* f 3 x (b 2 - x 2 ) 1/2 dx - 6 P x (a 2 - x 2 ) 1/2 dx + 6 f x (b 2 - x 2 ) 1/2 dx 

«-/ <-/ *-* it 

2(a 2 -x 2 ) 3 - 



2 (b 2 - x 2 ) 3 2 
3 



J n 



2 [b 2 - x 2 ) 3 2 
3 



„2\ 3 / 2 Iu3\ 3 / 2 



.2*3/2 



(b 2 -a 2 ) 0/ -(b 2 )°^ +f 0-(a 2 ) 



We calculate the mass geometrically: M = 6 A = S 



I 0-(b 2 -a 2 ) 



2x3/2 



fibi _ fa! _ Mb 3 -a 3 ) 
3 3 — 3 



M x ; 



-b 2 



( 7ra 2 \ 6tt < 
4 ) ~ 4 v 



6 ^ =^(b 2 -a 2 ). Thus,x=^ 



_ a (b 3 - a 3 ) 4 A ( \?-£ \ 

~ 3 ' <57r(b 2 -a 2 ) ~~ 3tt ^ b 2 - a 2 ^ 

tt _ Mx _ 4 (a 2 +ab+b 2 ) 
" M 3jr(a+b) 

n2 i .,2 i .,2 



4 (b - a) (a 2 + ab + b 2 ) _ 4 (a 2 + ab + b 2 ) 



37T (b - a)(b + a) 



3?r(a + b) 



; likewise 



(b) b h T a £ ( 3 ^T^) = (s) (***) = (£) (f) = f =* WW = (*,*) - *e limiting 

position of the centroid as b — > a. This is the centroid of a circle of radius a (and we note the two circles 
coincide when b = a). 



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Chapter 6 Additional and Advanced Exercises 423 



10. Since the area of the traingle is 36, the diagram may be 
labeled as shown at the right. The centroid of the triangle is 
(f , y). The shaded portion is 144 - 36 = 108. Write 
(x, y) for the centroid of the remaining region. The centroid 
of the whole square is obviously (6, 6). Think of the square 
as a sheet of uniform density, so that the centroid of the 
square is the average of the centroids of the two regions, 
weighted by area: 

36(f) + 108{x) _„ Je 36(f) +108(y) 



6 



and 6 



144 """ " — 144 

which we solve to get x = 8 — | and y = — 
x = 7 in. (Given). It follows that a = 9, whence y 



Set 



04 
9 




7 £ in. The distances of the centroid (x, y) from the other sides are easily computed. (Note that if we set y = 7 in. 



above, we will find x 



7 1 ) 

' 9-> 



11. 



2v/x" 



ds 



ldx 



JTv^ 



ldx= | [(1+x) 3 / 2 ] 







28 
3 



12. This surface is a triangle having a base of 27ra and a height of 27rak. Therefore the surface area is 

\ (27ra)(27rak) = 27r 2 a 2 k. 



2 



d£x 
dt- 



13. F = ma = t 
x = when t = 

n(\l 

W=fFdx = J o 

_ (12mh) 3 / 2 



a 



=> Ci 

(^mh) 1 /* 

12mh- v 12mh 



V 
X = 



dx 
dt 



F(t) - £ dt : 



Jo 



12m 
(12mh)V4 



3 m 

Then x 



C; v = when t = => C = =*> 



dx 
dt 



3 m 



h =*► t 



2 



dt 



(12mh) 1//4 . The work done is 

(12mh)V4 /ix r/, 

= Gib (12mh)«/ 4 



18n 



18m 



f -2^3mh 



V3mh 



t* 

12m 



•Ci; 



14. Converting to pounds and feet, 2 lb/in 



21b 12 in 
1 in 1 ft 



24 lb/ft. Thus, F = 24x 



W 



n 1/2 

/ 24x dx 

J() 



fl2x 



21 1/2 

Jo 



3 ft • lb. Since W = \ mvg - \ mvf , where W = 3 ft • lb, m = l-k lb 



v 32 ft/sec 2 / 



= 35q slugs, and Vi = ft/sec, we have 3 
s = - 16t 2 + v t (since s = at t = 0) => 

,32^ 



and the height is s = — 16 ( — 



vo(jf) 



= (&) 

~ 64 



'J- v 2 ) 

,320 *0) 



3-640 
64 



Vg = 3 • 640. For the projectile height, 
32t + v . At the top of the ball's path, v = = 
30 ft. 



hi 
32 



15. The submerged triangular plate is depicted in the figure 
at the right. The hypotenuse of the triangle has slope — 1 
=> y — (—2) = — (x — 0) => x = — (y + 2) is an equation 
of the hypotenuse. Using a typical horizontal strip, the fluid 

pressure isF = /(62.4).( ( gP i ).( 1 ^)dy 
= J_" 6 2 (62.4)(-y)[-(y + 2)] dy = 62.4 f~*(y 2 + 2y) dy 



62.4 | C + y 2 



(62.4) 



' 208 



-6 

32) 



( 62.4)[(-f+4)-(-f +36)] 



(62.4X112) 



2329.6 lb 




x--(y + 2) 



(4,-6) 



Copyright (c) 2006 Pearson Education 




424 Chapter 6 Applications of Definite Integrals 

16. Consider a rectangular plate of length I and width w. 
The length is parallel with the surface of the fluid of 
weight density u>. The force on one side of the plate is 

F = ujJ^ (-y)(£) dy = -U [£] ' = ^f- . The 

average force on one side of the plate is F ilv = ^ J (— y)dy 



if . Therefore the force ^f- 



-y 

-w 



(£w) = (the average pressure up and down) • (the area of the plate). 



one 
side 



17. (a) We establish a coordinate system as shown. A typical 
horizontal strip has: center of pressure: (x ."y ) 
= (|,y) , length: L(y) = b, width: dy, area: dA 
= b dy, pressure: dp — uj |y| dA = uib |y| dy 

=> F x = J y dp = J_ h y • ub |y| dy = -wb J y 2 dy 



-ujb 



-ub 



-h 3 

3 



-ubh 3 . 
3 



F = J dp = J_ h u |y| L(y) dy = -wb J_ h y dy 




b,-h) 



-o;b 



-wb 0- 



h 2 



^f . Thus, y = | 



2 J 



(b) A typical horizontal strip has length L(y). By similar 
triangles from the figure at the right, -4p = ~ y h ~ a 



L(y) 



jr (y + a). Thus, a typical strip has center 



of pressure: (x ,y ) = (x , y), length: L(y) 
= — j(y + a), width: dy, area: dA = - | (y + a) dy, 
pressure: dp = uj |y| dA = uj{— y) (— ^) (y + a) dy 



f (y 2 + ay) dy => F x = / y dp 



Ll w y f (y 2 + a y) d y = f J ,.,... (y :i + u V -ui> 



-(a+h) ' 
h 4 ' 3 



^'b 



P (y 3 

J -(a+h) W 



-(a+h) 



HO^H 1 



a + h) 4 a(a + h) 3 



)] 



uM | a 4 -(a + h) 4 
h 



-2h 
3 



-a(a + h) 3 



the distance below the surface is | h. 




(b,-a-h) 



12h 
^'b 
12h 



[3 (a 4 - (a 4 + 4a 3 h + 6a 2 h 2 + 4ah 3 + h 4 )) - 4 (a 4 - a (a 3 + 3a 2 h + 3ah 2 + h 3 ))] 
(12a 3 h + 12a 2 h 2 + 4ah 3 - 12a 3 h - 18a 2 h 2 - 12ah 3 - 3h 4 ) = f| (-6a 2 h 2 - 8ah 3 - 3h 4 



-ff (6a 2 + 8ah + 3h 2 ) ; F = / dp = / u |y | L(y) dy = f £^ h) (y 2 + ay) dy = f [£ + *£] 

i±* \( ^t. _|_ S 3 .^ _ f -(a + h) 3 , a(a + h) 2 Nl _ yb [~ (a + h) 3 -a 3 _,_ a 3 -a(a + h) 2 ] 

u* |" a 3 + 3a 2 h + 3ah 2 + h 3 -a 3 , a 3 - (a 3 + 2a 2 h + ah 2 ) 
h [ 3 + 2 

^ (6a 2 h + 6ah 2 + 2h 3 - 6a 2 h - 3ah 2 ) = ^ (3ah 2 + 2h 3 ) = ^ (3a + 2h). Thus, y = | 



■(a+h) 



3^2 

^ [2 (3a 2 h + 3ah 2 + h 3 ) - 3 (2a 2 h + ah 2 )] 



=^)(6a 2 + 8ah + 3h 2 ) _ , _y 
(^)(3a + 2h) ~~ V"2". 



( 6a 2 + 8ah + 3h 2 \ 
V_ 3a + 2h J 



the distance below the surface is 



6a 2 + 8ah + 3h 2 
6a + 4h 



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CHAPTER 7 TRANSCENDENTAL FUNCTIONS 



7.1 INVERSE FUNCTIONS AND THEIR DERIVATIVES 



1 . Yes one-to-one, the graph passes the horizontal test. 



2. Not one-to-one, the graph fails the horizontal test. 



3. Not one-to-one since (for example) the horizontal line y = 2 intersects the graph twice. 



4. Not one-to-one, the graph fails the horizontal test. 



5. Yes one-to-one, the graph passes the horizontal test 



6. Yes one-to-one, the graph passes the horizontal test 



7. Domain: < x < 1, Range: < y 




8. Domain: x < 1, Range: y > 
y 




y-f(x) 



9. Domain: -1 < x < 1, Range: - § < y < 




10. Domain: — oo < x < oo, Range: 
y 



y = f(x)/ 



-f<y<l 




11. The graph is symmetric about y = x. 



A 


V 




~""\ y=JT7 




\ 0<x<l 





1 



(b) y = \/\ -x 2 => y 2 = 1 - x 2 => x 2 = 1 - y 2 => x = y/l - y 2 => y = v 7 ! - x 2 = f _1 (x) 



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426 Chapter 7 Transcendental Functions 



12. The graph is symmetric about y = x. 




x = \ => y = i = f -1 tt 



13. Step 1: y = x 2 +l=^x 2 = y-l^x = y/y - 1 
Step 2: y = Vx - 1 = f _1 (x) 

14. Step 1: y = x 2 => x = — a/y, since x < 0. 
Step 2: y = - v /^=f- 1 (x) 



15. Step 1: y 
Step 2: y 



x 3 - 1 =>■ x 3 = y ■ 

VxTI = f-^x) 



(y+ 1)1/3 



16. Step 1: y = x 2 - 2x + 1 =>• y = (x - l) 2 
Step 2: y = 1 + yfx = f^x) 



=4- y/y = x — 1, since x > 1 =>- x=l + -/y 



17. Step 1: y = (x + l) 2 =4> ^/y = x+ 1, since x > -1 =>• x = y^y- 1 
Step 2: y= y^-l = f- 1 (x) 

18. Step 1: y = x 2 / 3 => x = y 3 / 2 
Step 2: y = x 3 / 2 = f- : (x) 

19. Step 1: y = x 5 => x = y 1 / 5 
Step 2: y = \/x = f ^(x); 
Domain and Range of f _1 : all reals; 

f(f-!(x)) = (x 1 / 5 ) 5 = x and f -1 (f(x)) = (x 5 ) 1/5 = x 



20. Step 1: y = x 4 



x = y 



1/4 



Step 2: y = S/x = f" 1 ^); 

Domain of f _1 : x > 0, Range of f -1 : y > 0; 

f(f-!(x)) = (x 1 / 4 ) 4 = xandf-^fW) = (x 4 ) 1/4 = x 



1/3 



21. Step 1: y = x 3 + 1 =>• x 3 = y - 1 => x = (y - 1) 
Step 2: y = V* - 1 = f _1 (x); 
Domain and Range of f _1 : all reals; 
f(f-!(x)) = ((x- l) 1 / 3 ) 3 + 1 =(x- 1)+ 1 =xandf- 1 (f(x))= ((x 3 + 1) - 1) 1/3 = (x 3 ) 1/3 = x 



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Section 7.1 Inverse Functions and Their Derivatives 427 



22. Stepl: y = |x — | =>• |x = y+| =4> x = 2y + 7 



Step 2: y = 2x + 7 = f -1 (x); 

Domain and Range of f _1 : all reals; 

f(f-i(x)) = I(2x + 7)-| = (x+Z) -| = xandf- 1 (f(x)) = 2(ix- |) + 7 = (x - 7) + 7 = x 



23. Step 1: y 
Step 2: y 



l 



f- J (x) 



Domain off 1 : x > 0, Range off 1 : y > 0; 
f(f- 1 (x)) = y^ = ^ = xandf-^fW) - ' 



,7^) 



x since x > 



24. Stepl: y = 4j^x 3 =i^x=^3 

Step 2: y= J 7 3 = yi = f-i(x); 
Domain of f _1 : x^O, Range of f _1 : y ^ 0; 



f(f- 1 (x)) 



(x-l/3)' 1 



i = x and f- J (f(x)) 



-1/3 



25. (a) y = 2x + 3 => 2x = y - 3 

=> X = | - | =* f-!(x) =|-| 
df | _ 2 df- 1 | ._ 1 



(C) 



dx I x=— 1 



dx 



(b) 



^y=/(x) = 2x + 3 




26. (a) y = \ x + 7 =4> 



y-7 



(c) 



=> x = 5y - 35 => f-\x) = 5x - 35 

df I _ 1 df^ I _ g 

dxL=-i 5' dx | x=34/5 



(b) 




27. (a) y = 5 - 4x =>■ 4x = 5 - y 

f-!(x) = 



5 _ y 

4 4 



5 _ i 
4 4 



(C) 



df I 

dx I x =l/2 



H ' dx 



(b) 




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428 Chapter 7 Transcendental Functions 

28. (a) y = 2x 2 



9 1 

x = y 



(c) 



— I 

dx I x=5 

df-' l 

dx 



y 2 ^ => f-^x) 

= 20, 



/2 

4x 



-4= X-V2 
2\/2 



J_ 
20 



(b) 




0.5 1 



29. (a) f(g(x)) = (\/x") 3 = x, g(f(x)) = Vx» = x 

(c) f (x) = 3x 2 => f (1) = 3, f (-1) = 3; 
g'(x)=ix- 2 / 3 => g'(l)=I, g '(-l) = I 

(d) The line y = is tangent to f(x) = x 3 at (0, 0); 
the line x = is tangent to g(x) = 3 x/x at (0, 0) 



(b) 




30. (a) h(k(x))=i((4x) 1 /3) 3 = x , 



k(h(x)) = (4 • fj 



1/3 



(c) h'(x) 



is 2 



h'(2) = 3, h'(-2) = 3; 



k'(x) = I (4x)- 2 / 3 => k'(2) = 1, k'(-2) = i 
(d) The line y = is tangent to h(x) = j at (0, 0); 
the line x = is tangent to k(x) = (4x) 1,/3 at 
(0,0) 



(b) 



3 

X J 

1"T 



y = (4x) 



1/3 



31. 



df 



3x 2 — 6x 



df- 1 

dx 



_ J_ 

— df 
= f(3) dx 



32. 



df 



2x-4 => 



df 1 

dx 



33. 



df- 



df- 1 
dx 



< = f(2) dx 



J- - 3 

(I) " 3 



34. 



dg- 1 I 



dg- 1 
dx 



_ J_ 

— dg 
lx = f(0) 37 



35. (a) y = mx =>• x = ± y => f x (x) = 5 x 

(b) The graph of y = f _1 (x) is a line through the origin with slope — . 

36. y = mx + b =>- x = — — — =>• f _1 (x) = — x — -; the graph of f _1 (x) is a line with slope — and y-intercept 



37. (a) y = x + 1 =>• x = y - 1 =>• f- a (x) = x - 1 

(b) y = x + b =>• x = y-b => f _1 (x) = x - b 

(c) Their graphs will be parallel to one another and lie on 
opposite sides of the line y = x equidistant from that 
line. 



'y = x 




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Section 7.1 Inverse Functions and Their Derivatives 429 



38. (a) y = -x+l => x = -y + 1 => f-\x) = 1 - x; 
the lines intersect at a right angle 

(b) y = -x + b => x=-y + b =>- f _1 (x) = b - x; 
the lines intersect at a right angle 

(c) Such a function is its own inverse. 




39. Let Xi ^ x 2 be two numbers in the domain of an increasing function f. Then, either Xj < x 2 or 
xi > X2 which implies f(xi) < f(x 2 ) or f(xi) > f(x 2 ), since f(x) is increasing. In either case, 
f( x i) 7^ f( x 2) and f is one-to-one. Similar arguments hold if f is decreasing. 



40. f(x) is increasing since x 2 > Xi 



I x „ + 5 > l,5.df_l 
3 x 2 "I" 6 -? 3 X! -|- g , dx — 3 



df" 1 _ J_ 
dx " (i) 



41. f(x) is increasing since x 2 > x : =>• 27X 3 , > 27xJ; y = 27x 3 => x = ± y 1 / 3 => f : (x) = | x 1/3 ; 



£ = 8 lx2 



df 



1 I _ _J_ _ 1 y -2/3 

dx "~ Six 2 I i x V3 ~~ 9x 2 / 3 — 9 A 



42. f(x) is decreasing since x 2 > Xj =^> 1 — 8xf < 1 - 8xJ; y = 1 - 8x 3 => x = \ (1 - y) 1/3 =>■ f ^x) = I (1 - x) 1/3 ; 



df 

dx 



-24x 2 



df- 1 1 -1 

dx " -24x 2 U(i-x)V3 6(1 -x) 2 / 3 



i(l-x)- 2 / 3 



43. f(x) is decreasing since x 2 > Xj =4> (1 - x 2 ) 3 < (1 - Xi) 3 ; y = (1 - x) 3 =^ x = 1 - y 1 / 3 => f : (x) = 1 



,1/3. 



-3(1 - x) 2 



df 1 

dx 



-3(1 -x) 2 



-1 

3x 2 3 



l x -2/3 



44. f(x) is increasing since x 2 > Xi => x 2 ' > x/ ; y = x 5 / 3 =>■ x = y 3 / 5 =>• f : (x) = x 3 / 5 ; 



df _ 5 Y 2/3 
dx — 3 A 



df 1 
dx 



1 

\ X 2 / 3 



3 _ 3 y-2/5 
5x 2/5 - 5 x 



45. The function g(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if Xi ^ x 2 then 
f( x i) 7^ f( x 2)> so — f( x i) 7^ — f( x 2) an d therefore g(xj) ^ g(x 2 ). Therefore g(x) is one-to-one as well. 

46. The function h(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if Xi ^ x 2 then 
f(xO ^ f(x 2 ), so ^ ± ^ , and therefore h(xi) ^ h(x 2 ). 

47. The composite is one-to-one also. The reasoning: If Xi ^ x 2 then g(xi) ^ g(x 2 ) because g is one-to-one. Since 
g( x i) 7^ g( x 2), we also have f(g(xi)) ^ f(g(x 2 )) because f is one-to-one. Thus, f o g is one-to-one because 

Xi ^ x 2 =* f(g( Xl )) / f(g(x 2 )). 

48. Yes, g must be one-to-one. If g were not one-to-one, there would exist numbers Xi ^ x 2 in the domain of g 
with g(xi) = g(x 2 ). For these numbers we would also have f(g(xi)) = f(g(x 2 )), contradicting the assumption 
that f o g is one-to-one. 



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430 Chapter 7 Transcendental Functions 



49. The first integral is the area between f(x) and the x-axis 
over a < x < b. The second integral is the area between 
f(x) and the y-axis for f(a) < y < f(b). The sum of the 
integrals is the area of the larger rectangle with corners 
at (0, 0), (b, 0), (b, f(b)) and (0, f(b)) minus the area of the 
smaller rectangle with vertices at (0, 0), (a, 0), (a, f(a)) and 
(0, f(a)). That is, the sum of the integrals is bf(b) — af(a). 



f(b) 




50. f'(x) = ^ cx + 1 — v*2 + = f^ — ^j. Thus if ad — be ^ 0, f'(x) is either always positive or always negative. Hence f(x) is 
either always increasing or always decreasing. If follows that f(x) is one-to-one if ad — be ^ 0. 

51. (go f)(x) = x => g(f(x)) = x => g'(f(x))f'(x) = 1 

52. W(a) = f\ [(f^Hy)) 2 - a 2 ] dy = = £ 2?rx[f(a) - f(x)] dx = S(a); W'(t) = ^[(f-^fCt))) 2 - a 2 ] f'(t) 

= 7T (t 2 - a 2 ) f '(t); also S(t) = 27rf(t)/ a ' x dx - 27r/' xf(x) dx = [7rf(t)t 2 - 7rf(t)a 2 ] - 27r/' xf(x) dx 
=> S'(t) = 7rt 2 f'(t) + 27rtf(t) - 7ra 2 f'(t) - 27rtf(t) = 7T (t 2 - a 2 ) f'(t) => W'(t) = S'(t). Therefore, W(t) = S(t) 
for all t G [a,b]. 



53-60. Example CAS commands: 
Maple : 

with( plots );#53 

f := x -> sqrt(3*x-2); 

domain := 2/3 ..4; 

xO := 3; 

Df := D(f); # (a) 

plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[l,3], legend=["y=f(x)","y=f '(x)"], 

title="#53(a) (Section 7.1)" ); 
ql := solve( y=f(x), x ); # (b) 

g:=unapply(ql,y ); 
ml := Df(x0); # (c) 

tl :=f(x0)+ml*(x-x0); 

y=tl; 

m2 := 1/Df(x0); # (d) 

t2 := g(f(x0)) + m2*(x-f(x0)); 

y=t2; 

domaing := map(f, domain); # (e) 

pi := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[l,9], thickness=[3,0] ): 

p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ): 

p3 := plot( tl, x=x0-l..x0+l, color=red, linestyle=4, thickness=0 ): 

p4 := plot( t2, x=f(x0)-l..f(x0)+l, color=blue, linestyle=7, thickness=l ): 

p5 := plot( [ [x0,f(x0)], [f(x0),x0] ], color=green ): 

display( [pl,p2,p3,p4,p5], scaling=constrained, title="#53(e) (Section 7.1)" ); 
Mathematica: (assigned function and values for a, b, and xO may vary) 

If a function requires the odd root of a negative number, begin by loading the RealOnly package that allows Mathematica 
to do this. See section 2.5 for details. 

«Miscellaneous "RealOnly" 

Clear[x, y] 

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Section 7.1 Inverse Functions and Their Derivatives 43 1 



{a,b} = {-2, 1};x0= 1/2; 

f[x_] = (3x + 2) / (2x - 11) 

Plot[{f[x],f[x]}, {x, a,b}] 

solx = Solve[y == f[x], x] 

g[yj = x/. solx[[l]] 

yO = f[xO] 

ftan[x_] = yO + f [xO] (x-xO) 

gtan[y_] = xO + 1/ f [xO] (y - yO) 

Plot[{f[x], ftan[x], g[x], gtan[x], Identity [x]},{x, a, b}, 

Epilog -+ Line[{{xO, yO},{yO, xO}}], PlotRange -> {{a,b},{a,b}}, AspectRatio 



Automatic] 



61-62. Example CAS commands: 
Maple : 

with( plots ); 

eq := cos(y) = x A (l/5); 

domain := .. 1; 

xO := 1/2; 

f := unapply( solve( eq, y ), x ); # (a) 

Df := D(f); 

plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[l,3], legend=["y=f(x)","y=f (x)"], 
title="#62(a) (Section 7.1)" ); 

ql := solve( eq, x ); # (b) 

g:=unapply(ql,y ); 

ml := Df(xO); # (c) 

tl :=f(xO)+ml*(x-xO); 

y=tl; 

m2 := 1/Df(x0); # (d) 

t2 := g(f(xO)) + m2*(x-f(x0)); 

Y=t2; 

domaing := map(f, domain); # (e) 

pi := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[l,9], thickness=[3,0] ): 

p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ): 

p3 := plot( tl, x=xO-l..xO+l, color=red, linestyle=4, thickness=0 ): 

p4 := plot( t2, x=f(xO)-l..f(xO)+l, color=blue, linestyle=7, thickness=l ): 

p5 := plot( [ [xO,f(xO)], [f(xO),xO] ], color=green ): 

display( [pl,p2,p3,p4,p5], scaling=constrained, title="#62(e) (Section 7.1)" ); 
Mathematica: (assigned function and values for a, b, and xO may vary) 

For problems 61 and 62, the code is just slightly altered. At times, different "parts" of solutions need to be used, as in the 
definitions of f[x] and g[y] 

Clearfx, y] 

{a,b) = {0, 1};x0=1/2; 

eqn = Cos[y] == x 1/5 

soly = Solve[eqn, y] 

f[x_] = y /. soly[[2]] 

Plot[{f[x], f[x]}, {x, a, b}] 

solx = Solve[eqn, x] 

g[yj = x/. solx[[l]] 

yO = f[xO] 

ftan[x_] = yO + f [xO] (x - xO) 

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432 Chapter 7 Transcendental Functions 

gtan[y_] = xO + 1/ f [xO] (y - yO) 

Plot[{f[x], ftan[x], g[x], gtan[x], Identity [x]},{x, a, b}, 

Epilog -» Line[{{xO, yO},{yO, xO}}], PlotRange -» {{a, b}, {a, b}}, AspectRatio ->• Automatic] 

7.2 NATURAL LOGARITHMS 



1. (a) In 0.75 = In \ = In 3 - In 4 = In 3 - In 2 2 = In 3 - 2 In 2 

(b) In | = In 4 - In 9 = In 2 2 - In 3 2 = 2 In 2 - 2 In 3 

(c) In \ = In 1 - In 2 = - In 2 (d) In \/9 = \ In 9 

(e) In 3v/2 = In 3 + In 2 1 / 2 = In 3 + \ In 2 

(f) In ^/Y15 = \ In 13.5 = ± In f = \ (In 3 3 - In 2) = ± (3 In 3 - In 2) 

2. (a) In jij = In 1 - 3 In 5 = -3 In 5 
(c) In lyfl = In 7 3 / 2 = | In 7 
(e) In 0.056 = In ^ = In 7 - In 5 3 = In 7 - 3 In 5 

In 35 + In 1 



I In 3 2 = | In 3 



(b) In 9.8 = In f = In 7 2 - In 5 = 2 In 7 - In 5 
(d) In 1225 = In 35 2 = 2 In 35 = 2 In 5 + 2 In 7 



(f) 



In 25 



In 5 + In 7 - In 7 _ 1 
2 In 5 2 



3. (a) In sin 9 — In 



' sin 8 \ 



In 



sin 



In 5 (b) ln(3x 2 -9x)+ln(i) = In (^*s) =ln(x-3) 



(c) i In (4t 4 ) - In 2 = In v^ - In 2 = In 2t 2 - In 2 = In (?f\ = In (t 2 ) 

4. (a) In sec + In cos 9 = In [(sec 6>)(cos 9)] = In 1 = 

(b) In (8x + 4) - In 2 2 = In (8x + 4) - In 4 = In (^f 1 ) = In (2x + 1) 

(c) 3 In Vt 2 - 1 - In (t + 1) = 3 In (t 2 - 1) 1/3 - In (t + 1) = 3 (±) In (t 2 - 1) - In (t + 1) = In ( (t + ( ,'f ,7 1} ) 

= ln(t-l) 



5. y = ln3x => y'=(i)(3)=i 
7. y = ln(t 2 ) => |=(i)(2t) = f 

9. y = ln l = i n3x -i => | = (j^) (-3x- 2 ) = - 1 

10. y = 1b 12 = In 10x-l =*£=(,£*) (-10x- 2 ) = - 1 

11. y = ki(0 + l) => I = (yi-f) (1) = ^ 12. y = ln(20 + 2) 



6. y = lnkx => y'= (i)(k) = x 

8 . y = in(t 3 / 2 ) => % = (&)am = i 



dt VtW \2 



dy = 

dfl V 29 + 2 



x,)(2) 



' + 1 



13. y = lnx 3 => I = (i) (3x 2 ) = | 14. y = (in x) 3 =* | = 3(ln x) 2 - £ (In x) = ^ 



Jy 



15. y = t(ln t) 2 =^ f = (In t) 2 + 2t(ln t) • ^ (In t) = (In t) 2 + ^fi = (In t) 2 + 2 In t 



16. y = tv/tat = t(ln t) 1 / 2 =» g = (In t) 1 / 2 + I t(ln t)' 1 / 2 - A (In t) = (In t) 1 ' 2 + !2e|L^ 
= (In t) 1 / 2 



2(ln t)V2 



17. y = ^lnx-^ =* S =x 3 lnx+ x 



1 4x 3 
4 x 16 



x 3 In x 



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Section 7.2 Natural Logarithms 433 



18. y = f In x - f => | = x 2 In x + ^ • i - ^ - v- 



x J In x 



iq v _ lnt . dy _ t(|)-(lnt)(l) _ t _ ln , 



on x, - 1 + lnt . dy _ t(j)-(l + lnt)(l) _ i _ i _ ln t _ tot 

ZU. y - ( =x> dt - (2 - (2 - ,2 



O, v _ Inx ^ .,,_ (l+lnx)(l)-(lnx)(l) 1 + '^-^ 

Z ' i - y l+lnx ^ > (1+lnx) 2 



(1+lnx) 2 x(l+lnx) 2 



09 „ - xlnx . ,J _ (l+lnx)(lnx + x-l)-(xlnx)(l) _ (l+lnx) 2 -lnx _ , lnx 

ZZ - y — 1+lnx ^ y - ,.,.-_« - 7i ,.-..« - I 



(1+lnx) 2 



(1+lnx) 2 



(1+lnx) 2 



23. y = ln(lnx) =>• y' 



« In x/\x7 xlnx 



24. y = ln(ln(lnx)) => y 



' — 1 . _d_ 
ln (ln x) dx 



(In (lnx)) 



1 1 d 



ln (ln x) ln x dx 



(lnx) 



l 



x (ln x) ln (ln x) 



25. y = #[sin (ln 9) + cos (ln &)]=>% = [sin (ln 9) + cos (ln 9)] + 6 [cos (ln 0)-\- sin (ln 9) ■ \] 
= sin (ln 9) + cos (ln 9) + cos (ln 9) - sin (ln 9) = 2 cos (ln 8) 



26. y = ln (sec 9 + tan 6) 



dy _ sec 8 



see 8 tan 6 + sec z 
sec 8 + tan 



sec 6>(tan + sec f 
tan 8 + sec # 



sec 8 



27. y 



xi/s+ 1 



lnx- i ln(x+ 1) 



1 / l * 

2 \\+\, 



2(x+l)-x 
2x(x+ 1) 



3x + 2 
2x(x+l) 



28. y=ilni^ = I[ln(l+x)-ln(l-x ) ] =» y' = \ [^ - (^) (-1)] = \ [fa 



-x+l+x 



1 



x)(l - x) 1 - X 2 



29. y 



1+lnt . dy (l-lnt)(j) -(l+lnt)(=l) 
1-lnt ^ dt (1-lnt) 2 



1 In t i 1 i In t 

t 1 ' t ^ ' 



(1-lnt) 2 t(l-lnt) 2 



30. y = Jin y/x = (in t V 2 ) 1/2 =* | = \ (in tV 2 ) ^ ■ | (in t V 2 ) = i (in t^ 2 ) ^ 2 - £ - - (t^) 



In 



t l/2\" 1 / 2 _ J^_ _ 1 -1/2 _ 1_ 



tl/2 2 



4tv/ln y't 



31. y = ln (sec (ln 8)) 



% = -^n» ■ f H (sec(ln 8)) = sec (ln 9) n ta " (ln g) • A (ln 0) = ^ 

d<9 sec (In 0) d# v v 7/ sec(ln#) 66 v 7 



32. y = ln ^^"g 9 = § (ln sin 9 + ln cos 8) - ln (1 + 2 ln 0) 



dy 1 / cos g _ sin g \ _ 9 

&B ~ 2 V sin 8 cos 9 J 1 + 2 ln ( 



cot 8 — tan i 



4 

(l + 21n£ 



33. y = ln 



(x 2 +l)-' 

(/I -X 



51n(x 2 + l)-iln(l-x) =* y' = ^ - \ (^) (-1) = jfc + ^ 



"■ v-ln A /^ = |[5 1n(x+l)-201n(x + 2)]^y' = l(^T-xf2) = I fefl^ 



5 3x + 2 



2 (x+l)(x + 2) 



35 ' ? = £„ '" <A«K => I = (m v^) ■ i (x 2 ) - (ln Jf) ■ £ (f ) = 2x In |x| - x In $ 



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434 Chapter 7 Transcendental Functions 

36. y = ff hit* =► | = (in \A) • £ (\^) - (in ^) • & (^ = (in \A) (|^ 3 ) - (in ^) (^ 



/^ 



In f/x In \/x 



3^2 2^ 



37. J* * i dx = [In |x|] Ig = In 2 - In 3 = In | 



39 - J ? ^dy = ln|y 2 -25|+C 



38. / ^ 5^2 dx = [In [3x - 2|] ° 1 = In 2 - In 5 = In § 

40 - /4^5dr = ln|4r 2 -5|+C 



41. /''j^h dt = [l n I 2 ~~ cos t[] o = In 3 - In 1 = In 3; or let u = 2 - cos t => du = sin t dt with t = 
=S> u = 1 and t = 7T => u = 3 =>• /" ^^ dt = J] 1 du = [ln |u|] \ = ln 3 - ln 1 = ln 3 



42. P \ttlie d0= M 1 -4cos6»|]o /3 = In [1 - 2| = -ln 3 = ln \\ or let u = 1 - 4 cos 9 => du = 4 sin 9 d<9 
with 61 = =>■ u = -3 and 9 = § =>• u = - 1 =>■ J" * y^f^g d0 = /_"' 1 du = [ln |u|] I* = - ln 3 = ln 1 



43. Let u = ln x => du = ~ dx; x = 1 =>■ u = and x = 2 =4> u = ln 2; 
£l_X^ dx = rj" 2 2u du = [ U 2]M = (ln 2) 2 

44. Let u = ln x =>• du = ~ dx; x = 2 =>• u = ln 2 and x = 4 => u = ln 4; 
£ n^ = IT u du = [ ln «] " = ln ( ln 4) - In (ln 2) = ln (HI) = ln (^f ) 

45. Let u = ln x => du = - dx; x = 2 =>• u = ln 2 and x = 4 => u = ln 4; 



ln(^)=ln2 



J» 4 n ln 4 , 

* = f u- 2 du = - i V 
2 x(lnx) 2 J ln2 L uJ ] 



1 , j_ _j_ 1 j_ 1 1 1 1 

In 4 ~r ln 2 ~~ ln 2 2 ~*~ ln 2 — 2 In 2 + ln 2 — 2 ln 2 — ln 4 



46. Let u = ln x => du = - dx; x = 2 => u = ln2 and x = 16 => u = ln 16; 



/; 



dx 



2x\/ln x 2 



f " '" u- 1 / 2 du = [u 1 / 2 ! '" '' = x/ln 16 - V/LT2 = x/4 In 2 - v/hT^ = 2v/ln2 - \f\nl = v/In2 

J 1„ 2 L J ln2 v v v v v v v 



47. Let u = 6 + 3 tan t =>• du = 3 sec 2 1 dt; 
/6^ dt =/T=Hu| + C = ln|6 + 3tant|+C 

48. Let u = 2 + sec y =>• du = sec y tan y dy; 

|f^ d y = /T=l n H+C = ln|2 + secy|+C 

49. Let u = cos | => du = - J sin | dx => -2 du = sin £ dx; x = => u = 1 and x = 5 => u = -4- 

2 2 2 2 2 yj2 



-Wft, 



tan f dx = I - — \ dx 

2 J() cos 2 



2 jC^T = h 2 ln M \'^ = ~ 2 ln 775 = 2 1° V 72 = l n 2 



p7T/2 /»: 

I tan | dx = I 

Jo 2 Jo 

50. Let u = sin t =4> du = cos t dt; t = f =>• u = -4= and t = ? =>■ u = 1 ; 

4 , /? 2 



J'" - / 2 r*V 2 r 1 1 

cottdt= / £?^dt= / ,-^ = [ln|u|] 
ir/4 Jtt/1 slnt J U\fi u ! 



-ln-L=ln^2 



51. Let u = sin | =>■ du = i cos f d0 =^ 6 du = 2 cos | d0; 9 = f => u = \ and = tt => u = ^ ; 



X> cot J d ^ = X/2 ? £r de = 6 / lA ^=6[ln|u|]^ = 6(lnf -lnl)=61nv^ = ln27 

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Section 7.2 Natural Logarithms 435 



52. Let u = cos 3x =>• du = —3 sin 3x dx => —2 du = 6 sin 3x dx; x = =>• u = 1 and x = -^ => u 



12 



/2' 



I 6 tan 3x dx = | 

Jo Jo 



6 sin 3x 
cos 3x 



dx 



-'/ 



du 
u 



-2[ln|u|] 



2 In 4- - In 1 = 2 In V 2 = In 2 

72 v 



53. 



f dx = r 

J 2\/x + 2x J 



dx ; let u = 1 + ,/x ^ du = -V dx; f dx 



2VTpTVJ) 
In |1 + Jx\ + C = In (1 + Jx) + C 



2v/x"(l + v^) 



/&=ln|u| + C 



54. Let u = sec x + tan x => du = (sec x tan x + sec 2 x) dx = (sec x)(tan x + sec x) dx =>- sec x dx = — ; 
secxdx . f_^_ = r (ln u) -i/2 . I du = 2 (ln u) 1 / 2 + C = 2^/ln(sec x + tan x) + C 

J uvlnu J u v 



Ay/ln (sec x + tan x) 



/ 



55. y = v/x(xTT) = (x(x + l)) 1 / 2 => lny = \ ln(x(x+ 1)) => 2 In y = ln(x) + ln(x + 1) 

=^ v' — ( 1 \ /vCv 4_ n ( 1 i M — V / "( x + 1 )( 2x + 1 ) _ 2x + l 
=► y - UJ V X ( X + 1) U + x+lJ - S(xTTj - 2Vx(x+l) 



X ' x+1 



56. y= VU 2 + 1)( X -D 2 => lny=i[ln(x 2 +l) + 21n(x-l)] =* L = 1 (^ + ^ 



y' = v/(x 2 + i) (x - i) 2 (^ + ^) = V(* 2 + i) (x - i) 2 [fenf^u] = ^vSt^ 



(2x 2 -x+l) x — 1 
1) 



57. y 



L_-f_t_^ ^lny=i[lnt-ln(t+l)] => J £ = I (I - ^ 



Y t+i vt+i/ 

dy _ l / t (1 ]_' 

dt 2 V t + 1 \ t t+1, 



1 / t 1 



2 V t+1 |_t(t+l)J 2v/t(t+ 1)3/2 



58. 



y = \/wh = W l + W~ 1/2 =* lny=|[lnt + ln(t+l)] 



1 dy 

y dt 



2 



Vt T t+ W 



dy 
dt 



1 / 1 2t + l 

2 V t(t+l) t(t+l) 



2t+l 
2 (t 2 + 1) 3/2 



59. y = y/6 + 3 (sin 0) = (0 + 3) 1 / 2 sin =^> In y = ± ln(0 + 3) + In (sin 9) => ± 



dv 



y d0 2(9 + 3) ' sin 



1 I cos ( 



dy 



\/6 + 3 (sin 0) [ 5?? ^ + cot0 



60. y = (tan 9) \/29 + 1 = (tan 0)(20 + 1) 1/2 =>■ In y = In (tan 0) + \ In (20 + 1) 



1 dy _ sec 2 
y d# — tan ( 



+ (§)G£t) 



dy 



(tan 0) y/29 + 1 (^| + ^) = (sec 2 0) ^/20T 



1 + 



tang 



1 dy 1 1 . 1 

y dt t ' t+1 ' t+2 



61. y = t(t + l)(t + 2) => In y = In t + In (t + 1) + In (t + 2) 

t(t + i)(t + 2) r < t+ i)(t+2)+Kt+2)+t(t+ 1) 



|=t(t+l)(t+2)(i 



t ' t+1 ' t+2/ 



t(t+l)(t + 2) 



3t 2 + 6t + 2 



62 - y = t(t+ixt + 2) =* In y = in 1 - In t - In (t + 1) - ln(t + 2) =» 

-^ dy _ 1 r_ 1 _ 1 _ 1 I _ 



y dt t t+1 t + 2 



i r_ i i i_] _ -l (t+i)(t+2)+t(t+2) + t(t+i) 

dt t(t+l)(t + 2) L t t+1 t + 2 J t(t+l)(t + 2) [ t(t+l)(t + 2) 

3t 2 + 6t + 2 



(t3 + 3t 2 + 2t) 2 



63. y 



In y = In (6 + 5) - In 9 - In (cos 9) 



1 dy _ _J_ _ 1 i sin£ 
y d8 ~ 9+5 ' cost 



S = (rK.)(jT3-i + '™») 



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436 Chapter 7 Transcendental Functions 



64. y 



=^ In y = In 6 + In (sin 9) - \ In (sec 0) 
cot - i tan (9) 



/secfl ' "" ' 2 

dy _ flsing /l 
d9 v^9 ^ 9 



1 dy _ 1 I cos 

y A9 ~ I 6 ~*~ sin 9 



fl (sec 9)(tan 9) 



2 sec ( 



65. y = f^ => lny = lnx+iln(x 2 +l)-fln(x+l) 



y' = ¥M\U 



(x+l) 2 / 3 x ~ x 2 +l 3(x+l) 



i_ _ 1 i x 2_ 

y x ~ x 2 + 1 3(x+ 1) 



(x+l) 11 



=> lny = i[101n(x+ l)-51n(2x+ 1)] => 



1-5 



""• y \/ (2x+l)= "^ "* } i 1 '""'^ ' ^ ■ ,iu ^" i ^J ->' y x + i 2x+l 



/ _ /(x+l) 10 / 5 



f_5 2_) 

V x + 1 2x + 1 1 



(2x+l) 5 Vx+1 2x + l 



67 - y= V^+i => lny = j[ lnx + ln ( x -2)- ln ( x2 + 1 )] => 



i _ 1 3 / x(x - 2) /l i 1_ _ 2x ' 

y — 3 V x 2 + 1 Vx + x-2 x 2 + 1 , 



t — I (l 4. _J_ _ 2x \ 

y 3U T i-2 x 2 + W 



68. y 



3 / x(x+l)(x-2) 

;x 2 + 1 ) (2x + 3) 



=4> In y = \ [In x + In (x + 1) + In (x - 2) - In (x 2 + 1) - In (2x + 3)] 



.,/ _ 1 3 / x(x+l)(x-2) /l , 1 . L_ 2x 



3 V (x 2 + l)(2x + 3) Vx r x + l ^ x-2 x 2 + 1 2x + 3; 



69. (a) f(x) = In (cos x) =^ f (x) 



sin X 

COS X 



tan x = =4> x = 0; f (x) > for - \ < x < and f (x) < for 



< x < | =>• there is a relative maximum at x = with f(0) = In (cos 0) = In 1 = 0; f (— |) = In (cos (— f )) 
= In I -4- ) = — j In 2 and f (|) = In (cos (|)) = In \ = — In 2. Therefore, the absolute minimum occurs at 

x = | with f (f) = — In 2 and the absolute maximum occurs at x = with f(0) = 0. 
(b) f(x) = cos (In x) =>• f'(x) = ~ sin x (lnx) =0 =4> x = 1; f'(x) > for \ < x < 1 and f'(x) < for 1 < x < 2 
=> there is a relative maximum at x = 1 with f(l) = cos (In 1) = cos 0= 1; f (I) = cos (in (|)) 
= cos (— In 2) = cos (In 2) and f(2) = cos (In 2). Therefore, the absolute minimum occurs at x = | and 

x = 2 with f (i) = f(2) = cos (In 2), and the absolute maximum occurs at x = 1 with f(l) = 1. 

70. (a) f(x) = x — In x => f'(x) = 1 — - ; if x > 1, then f'(x) > which means that f(x) is increasing 
(b) f(l) = 1 - In 1 = 1 =>■ f(x) = x - In x > 0, if x > 1 by part (a) =^ x > In x if x > 1 

71. J (In 2x - In x) dx = J (- In x + In 2 + In x) dx = (In 2) f dx = (In 2)(5 - 1) = In 2 4 = In 16 

72. A = f ° - tan x dx + f* \an x dx = f ° ^^ dx - f* " ^^ dx = fin Icos xll ° w , , - fin |cos xll „ /3 

J-jr/4 Jo J-tt/4 cosx Jo cosx LI IJ -jr/4 II IJ 

= (in 1 - In 4-) - (In | - In l) = In \fl + In 2 = | In 2 

73. V = nf^-^y dy = 4^ £ ^ dy = 4^ [In |y + l|]g = 4tt(1ii 4 - In 1) = 4^ In 4 

74. V = 7T P "cot x dx = 7T P ~ ^ dx = tt fin (sin x)l ^ = tt (in 1 - In ±) = tt In 2 

J tt/6 J tt/6 sin X 1 V 'J 7T/G V 2 / 

75. V = 2nf* x (i) dx = 2tt | J dx = 2tt [In |x|] j /2 = 2tt (in 2 - In ±) = 2tt(2 In 2) = tt In 2 4 = tt In 16 



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Section 7.2 Natural Logarithms 437 



76. V = 7T jY-^i— ) " dx = 21 tt £ dx = 27tt [In (x 3 +9)f = 277r(ln 36 - In 9) 
= 277r(ln 4 + In 9 - In 9) = 27tt In 4 = 54tt In 2 



77. ( i iM< = f-tax~l + (yT = l+(f-I)- = l+(%4) =(^) ^L=/ 4 7l + (y') 2 dx 
J^^p dx = J 4 8 (| + i) dx = [f + In |x|] = (8 + In 8) - (2 + In 4) = 6 + In 2 



(b) x=(^-21n(|) =* | = I-| => l + [£ 



(*)' 



i ( s — f i =i 



+(^) 2 =(^y 



8 + 2 In 3 = 8 + In 9 



12 



+ 21ny| =(9 + 2 In 12) -(1 + 2 In 4) 



78. L = Jj v/TT^i dx => -| = i =4> y = In |x| + C = In x + C since x>0 => = lnl + C =>• C = =^ y = lnx 



79. (a) M y = /;x(i)dx=l,M x = /;(i)(i)dx=ij; 2 idx=[-i] 2 =I,M=j; 2 I 



^dx= [lnlxllf =ln2 



X — M _ In 2 ~ i "** dHU J — M — In 2 ~ ^^ 



(b) 




80. (a) M y = f\ (-J-) dx = J/V/2 dx = f [ X 3/ 2 ] J 6 = 42; M, = /"(^ (-J.) dx = 1 /"i dx 
= 1 [In |x|]}° =ln4,M= /"^ dx = [2xW] \" = 6 => x = g = 7 and y = $ = ^ 

w ^ = r * fe) (^) dx = 4 /> = «>. m * = x 16 {&) (^) (^) dx = 2 /r x - 3/2 dx 

= -4[x-V 2 ]; c = 3,M = /;(-i ; )(^)dx = 4/;idx=[41n|x|]; 6 =41nl6=>x 



My _ J5_ 

M In 16 



and 



M 4 In 16 



1. ffl = 1 + I at (1, 3) => y = x + In |x| + C; y = 3 at x = 1 => C = 2 => y = x + In |x| + 2 



dx- 1 



dy 



tan x + C and 1 = tan + C =4> jj| = tan x + 1 =>■ y = J (tan x + 1) dx 



= In |sec x| + x + Ci and = In |sec 0| + + Ci => Ci = => y = In |sec x| + x 

83. (a) L(x) = f(0) + f (0) - x, and f(x) = In (1 + x) =>• f ' (x)| x=0 = -^ | x=o = 1 =>■ L(x) = In 1 + 1 • x => L(x) = x 
(b) Let f(x) = ln(x + 1). Since f"(x) = — l 2 < on [0, 0.1], the graph of f is concave down on this interval and the 

(x+lj 

largest error in the linear approximation will occur when x = 0.1. This error is 0.1 — ln(l.l) « 0.00469 to five 
decimal places. 



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438 Chapter 7 Transcendental Functions 



(c) The approximation y = x for In (1 + x) is best for smaller 
positive values of x; in particular for < x < 0.1 in the 
graph. As x increases, so does the error x — In (1 + x). 
From the graph an upper bound for the error is 
0.5 - ln(l + 0.5) « 0.095; i.e., |E(x)| < 0.095 for 
< x < 0.5. Note from the graph that 0.1 - In (1 + 0.1) 
w 0.00469 estimates the error in replacing In (1 + x) by 
x over < x < 0.1. This is consistent with the estimate 
given in part (b) above. 



0.5 








y = x/ 


0.4 








/ y' 


0.3 






/ , ■■' 


y = In (> 


0.2 




/' 






0.1 













O.'l 


Q'l 


013 


0!4 0:5 x 



84. For all positive values of x, 4- [ In | ] = i • — 53 = — j and ^ [ In a — In x ] = 0— 1 = — £ . Since In j| and In a — In x have 

x 

the same derivative, then In - = In a — In x + C for some constant C. Since this equation holds for all positve values of x, 
it must be true forx= 1 =4> In - =lnl — lnx + C = — lnx + C=>ln- = —In x + C. By part 3 we know that 
In - = -In x => C = =^ In - = In a - In x. 



85. y = In kx =>- y = In x + In k; thus the graph of 
y = In kx is the graph of y = In x shifted vertically 
by In k, k > 0. 




86. To turn the arches upside down we would use the 



formula y = — In jsin x| 



In 



10 15 20 




y ■ In I sin x| 



87. (a) 




(b) y' 



Since Isin xl and Icos xl are less than 



a+sm x 

or equal to 1, we have for a > 1 
^r < y' < -Xr for all x. 

a— 1 — J — a— 1 

Thus, lim y' = for all x ^> the graph of y looks 

a— >+oo 

more and more horizontal as a — > + oo. 



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(a) The graph of y = ^/x — In x appears to be concave 
upward for all x > 0. 



(b) y = Ay/x — In x => y' 



_j i 



1 

4x3/2 



Section 7.3 The Exponential Function 439 




+ i = i(-^ + i) 



rs ar x 



=> ,/x = 4 => x = 16. 



Thus, y" > if < x < 16 and y" < if x > 16 so a point of inflection exists at x = 16. The graph of 
y = y/x — In x closely resembles a straight line for x > 10 and it is impossible to discuss the point of 
inflection visually from the graph. 



7.3 THE EXPONENTIAL FUNCTION 

1. (a) e ln72 = 7.2 

2. (a) e ln ( x2 +y 2 )=x 2 + y 2 



(b) e- lnx2 -- 
(b) e- lnCO 



(c) e lnx ~ lny = e ln ' x / y ' = - 



3. (a) 21n v /e = 21ne 1 / 2 = (2)(i) lne= 1 
(c) lne(- x2 -y 2 > = (-x 2 - y 2 ) In e = -x 2 - y 2 

4. (a) In (e secfl ) = (sec 6»)(ln e) = sec 6 

(c) In (e 21nx ) = In (e lnx2 ) = In x 2 = 2 In x 



l j_ 

e ln0.3 03 



(c) e 



ln7rx— In 2 pln(7rx/2) yrx 



(b) In (In e e ) = ln(e In e) = In e = 1 



(b) In e( eX > = (e x ) (In e) = e x 



5. In y = 2t + 4 ^ e lny = e 2t + 4 ^> y = e 



a 2t+4 



6. In y = -t + 5 => e lny = e-'+ 5 => y = e~ t+5 



7. ln(y - 40) = 5t ^ e ln ( y - 40) = e 5t => y - 40 = e 5t => y = e 5t + 40 

8. ln(l - 2y) = t => e 1 "' 1 - 2 ^ = e l => 1 - 2y = e< => -2y = e< - 1 =>• y = - (^f 1 ) 



y-lx 



9. In (y - 1) - In 2 = x + In x =>• In (y - 1) - In 2 - In x = x =^ In v , x , ^ 

=>■ y - 1 = 2xe x => y = 2xe x + 1 

10. ln(y 2 - 1) -ln(y+ l) = ln(sinx) =^> In (y^j) = In (sin x) =>■ ln(y - 1) = In (sin x) =4- e 1 "^ 1 ) 

=^> y — 1 = sin x =>- y = sin x + 1 



y-i 



a In (sinx) 



11. (a) e 2k = 4 =>• In e 2k = In 4 => 2k In e = In 2 2 =^ 2k = 2 In 2 => k = In 2 

(b) 100e 10k = 200 =>■ e 10k = 2 => In e 10k = In 2 =>• 10k In e = In 2 => 10k = In 2 =* k = ^ 

(c) e k / 1000 = a => In e k / 100() = In a ^ ^ In e = In a =>■ j^ = In a =^ k = 1000 In a 

12. (a) e 5k = \ => In e 5k = In 4- 1 => 5k In e = - In 4 => 5k = - In 4 => k = - ^ 

(b) 80e k = 1 => e k = 80" 1 => In e k = In 80" 1 => k In e = - In 80 => k = - In 80 

(c) e (in0.8)k = 08 ^ ( e lll0 - 8 ) k = 0.8 => (0.8) k = 0.8 => k=l 



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440 Chapter 7 Transcendental Functions 

13. (a) e-°- 3t = 27 => In e -03 ' = In 3 3 => (-0.3t) In e = 3 In 3 => -0.3t = 3 In 3 => t=-101n3 

(b) e kt = 1 => In e kt = In 2- 1 = kt In e = - In 2 => t = - ^ 

(c) e (in0.2)t = 4 ^ (e 1 " 02 )' = 0.4 =S> 0.2 [ = 0.4 =4> In 0.2' = In 0.4 =>• t In 0.2 = In 0.4 => t = |^| 



14. (a) e- u - uu =1000 =>• In e- uult = In 1000 => (-O.Olt) In e = In 1000 => -O.Olt = In 1000 =*> t= -100 In 1000 
(b) e kt = i => In e kt = In 10" 1 = kt In e = - In 10 =» kt = - In 10 => t - ln 10 



10 
a (ln2)t _ 1 



=> e 



ln2\ 



2- y => 2 t = 2~ I =* t= -1 



(c) e^ 
15. e^ /t = x 2 => lnev /t = lnx 2 => \/t=21nx => t = 4(ln x) 2 



16. e x2 e 2x+1 =e l =>• e x2+2x+1 = e l 



^ In e x 2 +2x+i = Jn e t ^ t = x 2 + 2x + 1 



17. y = e~ 5x => y' = e~ 5x £ (-5x) => y' = -5e" 



18. y = e 2x / 3 => y' = e 2x / 3 £ (f ) 



y' = | e 2x / 3 



19. y = e 5 - 7x =>/ = e 5 ~ 7x £ (5 - 7x) => y' = -7e 5 - ?x 

20. y = e( 4 v /x "+ x2 ) =>. y = e (V*"+* 2 ) £ (4^ + x 2 ) => y' = (4- + 2x) e ( 4 v /x "+ x2 ) 

21. y = xe x — e x =4> y' = (e x + xe x ) — e x = xe x 

22. y = (1 + 2x)e~ 2x ^ y' = 2e~ 2x + (1 + 2x)e~ 2x £ (-2x) => y' = 2e~ 2x - 2(1 + 2x)e~ 2x = -4xe~ 2x 

23. y = (x 2 - 2x + 2) e x => y' = (2x - 2)e x + (x 2 - 2x + 2) e x = x 2 e x 

24. y = (9x 2 - 6x + 2) e 3x => y' = (18x - 6)e 3x + (9x 2 - 6x + 2) e 3x £ (3x) ^ y' = (18x - 6)e 3x + 3 (9x 2 - 6x + 2) e 3 

= 27x 2 e 3x 



25. y = e fl (sin 9 + cos 9) => y' = e"(sin 9 + cos 6) + e"(cos 9 - sin 6) = 2e e cos 9 

26. y = ln (39e-°) = ln 3 + ln 9 + ln e~ 8 = ln 3 + ln 6 - 9 => % = \ - 1 

27. y = cos (e-* 2 ) =* % = - sin (e^) A ( e -* 2 ) = (_ sin (V« 2 )) (e^ 2 ) ± (-6 2 ) = 29^ sin (e^) 



28. y = 9 3 e- w cos 56* => % = (3<9 2 ) (e" 26 cos 5(9) + (6 3 cos 56) t' 20 £ (-29) - 5(sin 59) (6 3 e- 20 ) 
= 9 2 e- 2e (3 cos 5(9 - 26 cos 59 - 56 sin 50) 



29. y = ln (3te~ l ) = ln 3 + ln t + ln e~ l = ln 3 + ln t - t 



dy _ 1 



1 



30. y = ln (2e _t sin t) = ln 2 + ln e _t + ln sin t = ln 2 - t + ln sin t 

cos t — sin t 



dy 
dt 



\ sin t / dt 



(sin t) = -1 



31. y = ln 



l+e" 



In e" - ln 1 + e? = 9 - ln l+e tf =>■ 



e\ -k ^y 



, l+e« 



l + e» =1 



l+e» 



1 

l+e« 



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Section 7.3 The Exponential Function 441 



32. y = ln^ =lnV0-ln(l + Vfl) 



dy 
J9 



(i + v / »)-\/e_ i i 

20 (l + v^) ~ 29(l + v^) ~ 29(1+91/2) 



33. y = e ( cost + lnt ) = e cost e lnt = te cost ^ % = e cost + te cost | (cos t) = (1 - t sin t) e cost 

34. y = e sint (In t 2 + 1) => § = e sint (cos t) (In t 2 + 1) + \ e sint = e sint [(In t 2 + 1) (cos t) + f] 

35. f * sin e l dt =>■ y' = (sin e lnx ) • A (In x) = sp 



36. y = /j^ In t dt => y' = (In e 2x ) • A (e 2x ) - (in e 4 v^) - A ( e VZ) = (2x ) (2e 2x ) - (4^) (e 4 v^) - A (4^) 

= 4xe 2x - 4 v /xe 4l A (-2=) = 4xe 2x - Se 4 ^ 

37. In y = e y sin x =^- I - ) y' = (y'e y ) (sin x) + e y cos x =4> y' I i — e y sin x) = e y cos x 



l ( 1 — ye y sin x \ 

1 \ y / 



e y cos x => y 



/ ye y cos x 

1 — ye y sin x 



38. lnxy = e x+y => In x + In y = e x+y => I + (i) y' = (1 + y') e x4 
=> y ' f i-y^ +i \ _ xe^+'-i _^ „/ _ y(xe*+>'-i) 



/( y -e x+y ) 



a x+y _ 1 



x(l -ye"+y) 



39. e 2x = sin(x + 3y) => 2e 2x = (1 + 3y') cos(x + 3y) => 1 + 3y' = ^f^ => 3y' = ^_ 



/ _ 2e 2x - cos (x + 3y) 
^ " 3cos(x + 3y) 

40. tan y = e x + In x =* (sec 2 y) y' = e x + 1 =>■ y' = (xeX + ^ cos2 y 

41. J(e 3x + 5e~ x ) dx = ^ - 5e~ x + C 42. J(2e x - 3e~ 2x ) dx = 2e x + § e~ 2x + C 

43. f e x dx = [e x ]J^ = e ln3 - e ln2 = 3-2=1 44. J_" n2 e~ x dx = [-e- x ]°_ ln2 = -e° + e lnI 

45. /8e( x+1 'dx = 8e( x+1 ' +C 46. J^e' 2 "- 1 ) dx = e^ 1 ' + C 

47. Jj^V 2 dx = [ 2eX/2 ] ta4 = 2 [ e( ' n9)/2 " e(ln4>/2 ] = 2 ( e ' n3 - e ' n2 ) = 2(3 - 2) = 2 

48. /* 16 e x / 4 dx = [4e x / 4 ] J" 6 = 4 ( e ( lnl6 >/ 4 - e°) = 4 (e ln2 - l) = 4(2 - 1) = 4 



1+2=1 



Let u = r 1 / 2 ^ du = i r" 1 / 2 dr => 2 du = r" 1 / 2 dr; 



J^ dr = / e rV2 • r- 1 / 2 dr = 2 J*e u du = 2e u + C = 2e fl/2 + C = 26^ + C 



50. Let u = -r 1 / 2 => du = - \ r~ 1/2 dr =4> -2 du = r" 1 / 2 dr; 

f^£ dr = JV rl/2 • r- 1 /' 2 dr = -2 Je 11 du = -2 e - r ' /2 + C = -2e"^ + C 



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442 Chapter 7 Transcendental Functions 

51. Letu=-t 2 => du = -2tdt => -du = 2t dt; 
J*2te- t2 dt = -Je u du = -e u + C = -e~ t2 + C 

52. Let u = t 4 =>• du = 4t 3 dt => \ du = t 3 dt; 
J" 1 3 e t4 dt = \ Je u du = \ e t4 + C 

53. Let u = - => du = - \ dx =4> -du = 4 dx; 

X X 2 X 2 ' 

J^ dx = J-e u du = -e u + C = -e'/ x + C 



54. Let u = -x 2 =4> du = 2x 3 dx => | du = x -3 dx; 



J"^£- dx = JV X 2 • x- 3 dx = 1 Je u du = \ e u + C = \ e~ x 2 + C = \ e" 1 /" 2 + C 



55. Let u = tan => du = sec 2 d0; = =>• u = 0, 9 = f =4> u=l; 

/^(l + e tanfl ) sec 2 d0 = J^ 'sec 2 d0 + /Je" du = [tan 0] l' A + [e u ] J = [tan (£ 
= (l-0) + (e-l) = e 



tan(0)] +(e : - e°) 



56. Let u = cot => du = - esc 2 d0; = | =>- u = 1, 

/^ 2 (1 + e c ° tfl ) esc 2 d0 = £'l esc 2 d0 - J" e u du 
= (0+l)-(l-e) = e 



| ^ u = 0; 



-cot0]:g-[e"];=[-cot(f)+cot(|)]-(e°-e 1 ) 



57. Let u = sec 7rt =>- du = 7r sec nt tan 7rt dt 



sec 7rt tan 7rt dt; 



J e sec (rt) sec (7rt) tan (7rt) dt = i Je u du = f + C = s-^ + C 

58. Let u = esc (7T + t) =» du = - esc (n + t) cot {n + t) dt; 

f e csc (7r+t) csc (7r + t) cot (7r + 1 ) dt = - Je u du = -e u + C = -e csc fr+Q + C 

59. Let u = e v => du = e v dv => 2 du = 2e v dv; v = In | => u = |, v = In f =>• u = §; 

J»ln (7r/2) r^/^ /O 

i„(,/6, 2e v cose v dv = 2j V6 cosudu=[2sinu]^ = 2[sin(|)-sin(|)]=2(l-I) = l 



60. Let u = e x => du = 2xe x dx; x = =4> u = 1, x = V In 7r =>• u = e ln7r = 7r; 

J 2xe x cos I e x ) dx = J cos u du = [sin u] * = sin(7r) — sin(l) = — sin(l) w —0.84147 



61. Let u = 1 + e r =>• du = e r dr; 

JlT? dr = Ji du = In |u| + C = In (1 + e r ) + C 

62. f T -h r dx= f^ydx; 

J 1 + e 1 J e x + 1 ' 

let u = e~ x + 1 =>■ du = — e~ x dx => — du = e _x dx; 
J^TT dx = - / i du = - In |u| + C = - In (e- x + 1) + C 



63. 



Jy 



e' sin (e ( - 2) =4> y = JV sin (e l - 2) dt; 



let u = e l - 2 =>• du = e' dt =4> y = J sin u du = - cos u + C = - cos (e l - 2) + C; y(ln 2) = 

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Section 7.3 The Exponential Function 443 

=> - cos (e ln2 - 2) + C = => - cos (2 - 2) + C = => C = cos = 1; thus, y = 1 - cos (e l - 2) 



64. 



dy 



e ' sec 2 (7re ') => y = I e ' sec 2 (7re ') dt; 



let u = 7re ' =>• du = -7re 'dt =4> - ^ du = e ' dt =>• y = — - / sec 2 u du = - - tan u + C 

TT J TT J TT 

= - I tan (tkT 1 ) + C; y(ln 4) = § =* - i tan (™-' n4 ) + C = 2 => - ± tan (tt • i) + C = £ 
=> -±(1) + C= ^ => C= ^;thus, y = 2 - I tan^e"') 



65. S = 2e~ x ^ f = -2e~ x + C; x = and ^ = => 0= -2e° + C => C = 2; thus & = -2e- x + 2 

ax 2 dx ' dx ' dx 

=> y = 2e~ x + 2x + Ci; x = and y = 1 =^> 1 = 2e° + Ci => Ci = -1 => y = 2e~ x + 2x - 1 = 2 (e~ x + x) - 1 



66. 



d 2 v 
dt- 

|=t-ie-+^-l ::- V 



I - c- 1 -> ^ = t ~ 5 e 2t + C; t = 1 and f = => = 1 - | e 2 + C => C = | e 2 - 1; thus 



a 2t _^ dy 
dt 

e 2 



it 2 -ie 2 <+(ie 2 



l) t + Ci;t = 1 andy 



i _ I ».2 i 1 -2 

2 4 C " T " 2 



Ci = -!- ip 2 

v-1 2 4 



y=it 2 -Ie 2t - 



'1 a 2 



e 2 -l)t-(i + Ie 2 ) 



67. f(x) = e x - 2x =4> f (x) = e x - 2; f (x) = =>■ e x = 2 => x = In 2; f(0) = 1, the absolute maximum; 

f(ln 2) = 2 — 2 In 2 w 0.613706, the absolute minimum; f(l) = e — 2 « 0.71828, a relative or local maximum 
since f "(x) = e x is always positive. 

68. The function f(x) = 2e sln ' x /^ has a maximum whenever sin | = 1 and a minimum whenever sin | = — 1. 
Therefore the maximums occur at x = 7r + 2k(27r) and the minimums occur at x = 3tt + 2k(27r), where k is any 
integer. The maximum is 2e ss 5.43656 and the minimum is - w 0.73576. 

69. f(x) = x 2 In i =4> f'(x) = 2x In i + x 2 (\\ (-x~ 2 ) = 2x In i - x = -x(2 In x + 1); f'(x) = => x = or 

In x = — i . Since x = is not in the domain of f, x = e~ 1//2 = -4= . Also, f '(x) > for < x < -4- and 

z v e v e 

f '(x) < for x > -j- . Therefore, f I -K- ) = \ In \fk = \ In e 1 / 2 = ^ In e = i is the absolute maximum value 



of f assumed at x 



70. f(x) = (x - 3) 2 e =* f (x) = 2(x - 3) e x + (x - 3) 2 e x 
= (x - 3) e x (2 + x - 3) = (x - l)(x - 3) e x ; thus 
f (x) > for x < 1 or x > 3, and f'(x) < for 
1 < x < 3 => f(l) = 4e w 10.87 is a local maximum and 
f(3) = is a local minimum. Since f(x) > for all x, 
f(3) = is also an absolute minimum. 




r( x ) = (x-l)(x-3)< 



71. X"V x -e x )dx=[^ -e x ]" 3 =(^-e-) - (f - e°) = (| - 3) - (I - 1) = §-2 = 2 



72. J o 2 '° 2 (e x / 2 - e- x / 2 ) dx = [2e x / 2 + 2e- x / 2 ] 



2 In 2 



2e ln2 + 2e- ln2 j - (2e° + 2e u ) = (4+ 1) - (2 + 2) = 5 - 4 = 1 



73 - L = Xv T +? dx 



I = ?|2 => y = e x / 2 + C; y(0) = => = e° + C => C = -l => y = e x / 2 - 1 



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444 Chapter 7 Transcendental Functions 



74. S = 2tt f 

Jo 

p In 

= 2tt 

Jo 



' e' - + e~ 



ln2 /- e y + e ->\ 2 



1 + \ (e 2 y - 2 + e- 2 y) dy 
dy= | J o " ,2 (e 2 5' + 2 + e- 2 >')dy 



■n [I P.2y 

2 L2 



e'> + 2y 



_ 1 e -2yl ln2 

2 C Jo 



|(i. 4 + 21n2-I-i) 



= |[(Ie 2 '« 2 + 21n2-Ie- 2 >« 2 )-(i+0-i)] 
|(2-I+21n2)=7r(if+ln2) 



75. (a) -f (x ln x - x + C) = x • \ + ln x - 1 + = ln x 



(b) average value = ^y I ln x dx = ^rj [x ln x — x] *j = ~j [(e ln e — e) — (1 ln 1 — 1)] 
= J- r (e-e+l)=^- r 

e— 1 v / e — 1 



76. average value = ^y I - 



dx = fin Ixll t = ln 2 — ln 1 = ln 2 



77. (a) f(x) = e => f (x) = e x ; L(x) = f(0) + f'(0)(x - 0) =>• L(x) = 1 



(b) f(0) = 1 and L(0) = 1 =>• error = 0; f(0.2) = e (u 

(c) Since y" = e x > 0, the tangent line 
approximation always lies below the curve y = e x 
Thus L(x) = x + 1 never overestimates e x . 



1.22140 and L(0.2) = 1.2 
y 



error w 0.02140 



y = e 

I 
I 




78. (a) e x e~ x = e( x - x ) = e° = 1 =$■ e" x = h for all x 



• 2*i = p x i(J- 



x l p— x 2 — f> x l~ x 2 



e A1 e 



(b) y = (e Xl )" 2 =>• ln y = x 2 ln e Xl = x 2 Xi = XiX 2 => e 



In y — o x l x 2 



X A X 2 _ „ x l x 2 



(e Xl ) 



79. f(x) = ln(x) - 1 => f (x) 



X n +1 



ln (x„) - 1 



x n+ i = x n [2 — In (x n )] . Then Xi = 2 



=4> x 2 = 2.61370564, x 3 = 2.71624393 and x 5 = 2.71828183, where we have used Newton's method. 

80. e lnx = x and ln (e x ) = x for all x > 

81. Note that y = ln x and e y = x are the same curve; / ln x dx = area under the curve between 1 and a; 

pin a 

I e y dy = area to the left of the curve between and ln a. The sum of these areas is equal to the area of the rectangle 
=4> I ln x dx + I e y dy = a ln a. 

82. (a) y = e x =>■ y" = e x > for all x =4> the graph of y = e x is always concave upward 

/>lnb 

(b) area of the trapezoid ABCD < J e x dx < area of the trapezoid AEFD => ± (AB + CD)(ln b - ln a) 

< J e x dx < ( e "'" + e '" b j (ln b - ln a). Now \ (AB + CD) is the height of the midpoint 

M = e ( llla+lnb '/ 2 since the curve containing the points B and C is linear =>• e (lna+lnb ^ 2 (ln b — ln a) 

< £" b e x dx < ( e "'" + e "' Ll ) (ln b - ln a) 

f hlb tab 

(c) I e x dx = [e x ] ° a = e lnb — e lna = b — a, so part (b) implies that 

e (ina+inb)/2 (i n b - In a)< b - a < ( e '" + e ' nb ) (ln b - ln a) => e ( lna+lnb '/ 2 < ln !g:f na < S±* 

Vab < 



„ln a/2 „lnb/2 ^ b-a , a + b 

C ' C ^ ln b - ln a ^ 2 



/pin a ,/plnb ^ b-a ~ a + b 

C V C ^ ln b - ln a ^ 2 



b — a ^- a + b 
lnb-lna ^ 2 



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Section 7.4 a x and log a x 445 



7.4 a" and log a x 

1. (a) 5 l0&7 = 7 (b) 8 l0& 

(d) log 4 16 = log 4 4 2 = 2 log 4 4 = 2-1=2 
(f) log 4 (1) =log 4 4- 1 = -llog 4 4 = -l-l = -l 



* = Ji 



(c) 1.3 log '-' 75 = 75 



(e) log 3 y/3 = log 3 3 1 ' 2 = | log 3 3 = 4-1 



0.5 



2. (a) 2 l0fc3 = 3 (b) 10 loe '°( I/2 » = \ 

(d) log,, 121 = log,, 1 1 2 = 2 log,, 11=2-1=2 

(e) log 121 11 = log, 2 , 121 1 / 2 = (I) log 121 121 = (I) - 1 = \ 

(f) log 3 (i) = log 3 3- 2 = -2 log 3 3 = -2 - 1 = -2 

3. (a) Let z = log 4 x => 4 Z = x => 2 2z = x =4> (2 Z ) 2 = x =>■ 2 Z = v /x 

(b) Let z = log 3 x =4> 3 Z = x => (3 Z ) 2 = x 2 =^ 3 2z = x 2 => 9 Z = x 2 

(c) log, (e('° 2 ) si " x ) = log 2 2 si,,x = sinx 

4. (a) Let z = log 5 (3x 2 ) =>■ 5 Z = 3x 2 => 25 z = 9x 4 

(b) log, (e x ) = x 

(c) log 4 (2 e * sinx ) = log 4 4< e " sinx )/ 2 = ^p 



( C ) 7T k,g ' 7 = 7 



c /„\ log2 x lnx _^_ ln_x lnx ln_3 ln_3 

J ' W log 3 x ~~ In 2 ■ In 3 ~~ In 2 " lnx ~~ In 2 

( r \ logx a In a . In a In a . In x 2 2 In x ^ 

l°g x 2 a In x ' In x 2 In x In a In x 



f> ('a , l IPgi x lnx _^_ lnx lnx In 3 1 

°- W log 3 x ~~ In 9 ■ In 3 ~~ 2 In 3 ' lnx ~~ 2 

(U) l0 ^' = fax ^_ lnx lnx (I) ln2 Jn2_ 

W logyjx i n /Jo • to ,/2 (l)lnio' lnx ln 10 



(h\ log2 x l2_x _;_ lnx _ lnx ln_8 3 ln 2 _ o 

W logg x — ln 2 • ln 8 — ln 2 " ln x — ln 2 — J 



(C) 



log a b _ lnb _^ lnji _ lnb _ lnb _ / ln b \ 2 
logb a ln a ' ln b ln a ln a V ln a / 



7. 3 log1 (7) + 2 log2 ( 5 > = 5 log5 W =>. 7 + 5 = x => x = 12 

8. 8 logs (3) - e ln5 = x 2 - 7 log7 < 3x > => 3 - 5 = x 2 - 3x => = x 2 - 3x + 2 = (x - l)(x - 2) => x = 1 or x = 2 

9. 3 log1 < x2) = 5e lnx - 3 • 10 loglu ^ => x 2 = 5x - 6 => x 2 - 5x + 6 = => (x - 2)(x - 3) = => x = 2orx = 3 



10. ln e + 4- 21og " W = I i O g 10 100 => 1 + 4 log ^ x 2 > = i log 10 10 2 =^ 1 + x" 

=> x 2 - 2x + 1 = => (x - l) 2 = => x = 1 

11. y = 2" =>• y' = 2 s ln 2 12. y = 3" x 

13. y = 5^ s =» | = 5^ (In 5) (1 s- 1 /' 2 ) = (|l|) 5^ 

14. y = 2 s2 => f = 2 s2 (ln 2)2s = (ln 2 2 ) (s2 s2 ) = (ln 4)s2 s2 



(2) =► 1 + A - f = 



y ' = 3- x (ln3)(-l) = -3" x in3 



15. y = x^ =>• y' = 7TX 



' = TTY^-') 



16. y = t'- e =>• 



i=d-0r 



17. y = (cos 0)^ 



dv 



V2 (cos 6»)( v5 "') (sin 6») 



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446 Chapter 7 Transcendental Functions 
18. y = (In &=>% = 7r(ln 9)^ (I) = ^f^ 



19. y = 7 secS In 7 



20. y = 3 lme In 3 



dy 



dy 



(7 secS In 7) (In 7)(sec (9 tan 9) = 7 secS (ln 7) 2 (sec 6 tan 0) 



(S"" 1 " In 3)(ln 3) sec 2 9 = 3 ,imS (ln 3) 2 sec 2 



21. y = 2 sin31 => | = (2 si ° 3 ' ln2)(cos3t)(3) = (3cos3t)(2 sin3l )(ln2) 

22. y = 5- cos2 ' => f = (5- cos2l ln5)(sin2t)(2) = (2sin2t)(5- cos2, )(ln5) 

23. y = log 2 50 = |f =* S = (&) (i) (5) = lA* 



24. y = log_, (1 + 01n3) 



ln(l+flln3) 
In 3 



dy 



unMl+kl)^) 



1 + In 3 



25 v = — + — 

Z - J - y In 4 T In 



In x i In xf In x i ^ In x o In x ^ _J 3 

4 ~~ In 4 "^ Z In 4 — J In 4 =? " ~~ x In 4 

x In e In x x In x 



2 6- y = m - S = ±5 - ^ = (21b) (* - ln *) => y' = (dr?) (1 - i) 



27. y = log 2 r.log 4 r= (£§) (£f) 



28. y = log 3 r.log 9 r=(^)(^) 





ln 2 r 




=>• 


dy 


(lr 


2)(ln 


4) 


dr 




ln 2 r 




=>■ 


dy 


(li 


3)(ln 


9) 


dr 



[(I^k4j]( 21nr )(7) 



x- 1 
2xln5 



21nr 
r(ln 2)(ln 4) 



29- y = bfc((HS) M ) 



WHi) 



(In 3) In ( 



In 3 
-2 



In 3 



^Z = [ 1 1 (2 In r) CM = 21nr 

dr [ (In 3)(ln 9) J ^ ln l ' \t) r(ln 3)(ln < 



ln(|±i) =ln(x+ l)-ln(x- 1) 



dy _ 1 _ 1 

dx — x+1 x-1 ~~ (x+l)(x-l) 



30- y = log 5 V(3TT2) = 10 ^ (5TT 



I I tx \(ln5)/2 

1„5 , , 7x ^(to5)/2 _ In [jZfry 



In 5 



:¥) 



i?Mi) 



= | In 7x - \ In (3x + 2) ^ 
31. y = sin(log 7 0) = sin (£|) 



Jv 



J _ (3x + 2)-3x 



dx 2-7x 2-(3x + 2) ' 2x(3x + 2) x(3x + 2) 



% = M&) 



- [cos 



(hi)} f_L_ 

\ In 77 J Vein 7 



= sin 



1 In MM 

2 m V 3x + 2 J 



(log, 0) + jpj cos (log, 



0) 



32. y = log, (^f^) 



in#cos#\ _ In (sin g) + In (cos g) - In e" - In 2" _ In (sin g) + In (cos t 



I In 2 



In 7 



ss = r ■ w 7. " r s ivi ,i - v^i - rl = (rM (cot - tan 9 - 1 - In 2) 

dy (sin 0)(ln 7) (cos 0)(ln 7) In 7 In 7 V In 7 / v ' 



33. y = log 5 e*=^ = ^ => y'- 



34. y = log 2 



f x 2 e 2 \ _ In x 2 + In t 

V2 % /xTiy " 



In 5 



- In 2 - In %/x + 1 21nx + 2-ln2-lln(x+l) 



In 2 
4(x+l)-x 



In 2 



3x + 4 



^ y xln2 2(ln2)(x+l) 2x(x+l)(ln2) 2x(x+l)ln2 

35. y = 3 l0fcl = S^'VC" 2 ) => g = [3 (tal V('» 2 )(ln 3)] (^ = 1 (log 2 3) 3 lusl 



36. y = 3 log 8 (log, t) 

_ i 



3 1n(log 2 t) _ 3 In (^) dy _ / J_ 



In 8 



dt V In 8 I \ (In t)/(ln 2) V t In 2 I t(ln t)(ln 8) 



t(ln t)(ln 2) 



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Section 7.4 a x and log a x 447 



37. y=l0g 2 (8t'° 2 ) = 1 " 8 + W t '" 2 ) = 31n2 + (ln2)(lnt) = 3 + ln t _ dy _ 1 



In 2 



ln2 



dt t 



tin 



((e'"T° 



38- Y = ^3^ = ^ = ^^T 1 = tsint =* f = sint + tcost 



39. y = (x + l) x => ln y = ln (x + l) x = x ln (x + 1) =>• y - = ln (x + 1) + x • ^-^ => y' = (x + l) x [^ + ln (x + 1) 

40. y = x< x+1 > => ln y = ln x< x+1 > = (x + 1) lnx =>■ y - = ln x + (x + 1) (±) = ln x + 1 + \ 

=> y> = x (x+1 » (1 + i+lnx) 

41. y= (0)'= (t 1 / 2 )'^ 2 => lny = lnt'/ 2 = (±) ln t => i$ = (§)(lnt)+'^ '^ - lnt ' : 



v2y Vt/ 2 '2 



l = (^)'(1 t + i) 



42. y = t^ = t< ll/2 ) => ln y = ln t^ = (t 1 / 2 ) (ln t) => 1 | = (1 I" 1 / 2 ) (i n t) + t 1 / 2 (I) = ^ 



dt 



f lnt+2 \ tN /t 
V 2\A / 



43. y = (sin x) x =>• ln y = ln (sin x) x = x ln (sin x) =4> - = ln(sin x) + x (^S 5 ^) =4> y' = (sin x) x [ln (sin x) + x cot x] 



44. y = x s,nx =>• ln y = ln x smx = (sin x)(ln x) =>• - = (cos x)(ln x) + (sin x) 

. / _ sinx f s in x + x(ln x)(cos x) 



' X \ sin x + x (ln x)(cos x) 



45. y = x lnx , x > => ln y = (ln x) 2 => y j = 2(ln x) Q) => y' = (x" lx ) ( ] -^f) 



46. y = (ln x) lnx => ln y = (ln x) ln (ln x) 
=>?= ( ln(ln x x) + 1 )(lnx)'" x 



;i)ln(lnx) + Onx)(^)i(lnx) 



ln(lnx) , 1 



47- J*5 x dx=^+C 
49 - fo**M=fo{k)'M 



48. J(1.3) x 



QX — In (1.3) +( - 



M In 



\ n (l) 2(lnl-ln2) 2 ln 2 



(*0 n0 a 

5o - Jj- 9 m=L& M 



In In 



TIT (1-25) 



-24 _ 24 



In 1 - ln 5 ~~ ln 5 



51. Let u = x 2 => du = 2x dx =>■ | du = x dx; x=l =>• u=l,x= \J 2 =>■ u = 2; 



/ ^ x2) d X =/;(i)2"du=H^]? = (^2)(2 2 -2 



l 

In 2 



52. Let u = x 1 / 2 ^> du = i x" 1 / 2 dx => 2 du 



dx . 



;x=l =4> u=l,x = 4 =4> u = 2; 

2 



r $ dx = r 2xi/2 • *- i/2 dx = 2 r 2u du = i 2 ^] x = (A) ( 23 - 2 2 ) 



4 
ln 2 



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448 Chapter 7 Transcendental Functions 



53. Let u = cos t =^ du = - sin t dt => - du = sin t dt; t = =>- u = 1, t = 5 => u = 0; 



6 

In 7 



£ /2 7"« sin t dt = -X° 7" du = [- £] J = (-1) (7° - 7) 
54. Let u = tan t => du = sec 2 1 dt; t = =>■ u = 0, t = t => u=l 

/; /4 (i) imi sec 2 tdt=j; ) a) u du= 



In 3 J V 3 J V 3 J — 3 In 3 



55. Let u = x 2x => In u = 2x In x => ± g = 2 In x + (2x) Q) => jjs = 2u(ln x + 1) =>■ \ du = x 2x (l + In x) dx; 
x = 2 => u = 2 4 = 16, x = 4 => u = 4 8 = 65,536; 



J»4 r»65,536 ,_ „, 

x*(l + In x) dx = I J du = 1 [u] i6 - = | (65,536 - 16) 



56. Let u = In x => du = * dx; x = 1 =>• u = 0, x = 2 =4> u = In 2; 

1 \ ol»2 _ O0\ _ 2'° 2 -l 



65,520 
2 



32,760 






In 2 



58. Jx( v/5 " 1 )dx= 4j+C 
59. /; (^2 + l) x^ dx = [x(^)] * = 3 (^) 60. f>^ <*=[£] I = *fe^ = f=* = £ 



61 . / ^ d* = /(£*>) (I) dx; [u = In x=> du= I dx ] 

- J(n^)(Ddx= In ^/udu=( i ^)(Iu 2 ) + C=^ + C 

62. J] ^ dx = J] (|§) (i) dx; [u = In x => du = ± dx; x = 1 => u = 0, x = 4 => u = In 4] 

-r(^)(Ddx = r\^)udu=(^)[Iu 2 ]r=(^)[I(ln4) 2 ]=f^ = ^ = ln4 

63- J^ 3 ^ dx = //(i^) (£f ) dx = /> dx = [I (In x) 2 ] J = I [(In 4) 2 - (In I) 2 ] = \ (In 4) 2 
= \ (2 In 2) 2 = 2(ln 2) 2 

64. j; e 21nl0 f s '" x) dx = p^fiP (i) dx = [(In x) 2 ]; = (In e) 2 - (In I) 2 = 1 



65 ■ £^M^ ** = & T Mx + 2)](^) dx = (£) [filfe±ae] * = (£)[ 



J_\ (In 4) 2 _ (In 2)" 
vln 2/ 2 2 



1 \ r4(ln2) 2 (ln2) 2 1 _ 3 



In 2 



/'10 
1/10 



login (lOx) ^„ _ 10 



dx 



10 \ [4(ln 10) 2 



In 10 



X / Jln(10x)](^)dx=( i ^)[M^)!] i 



to 

/10 



' 10 \ (In 100) 2 _ (In 1)' 
vln 10/ 20 2 



vln 10/ 20 



2 In 10 



67 . r;2 j£f ^ dx = _ 2 _r; ln(x + 1)( _L T)dx _. , , r^ n,r_.^,[^_.^ 

= ln 10 



68. /;^^>dx=^X 3 ln( X -l)(^ 



l) dx - Vln; 



2 \ | (ln(x-l)) 2 



2 \ I (In 2) 2 (In l) 2 



In II 2 2 



In 2 



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Section 7.4 a x and log a x 449 



69 - I^t-, = I (%r) Q) dx = d" 10) / (£) (i) dx; [u = tax => du = 1 dx] 

-» (Inl0)/( I ^) (i) dx = (tal0) / ±du = (lnl0)ln|u|+C = (lnl0)ln|lnx|+C 

70. f -7^2 = f -£ti = On 8) 2 f 2^ dx = (In 8) 2 S 1 ^ + C = - ^ + C 

J x(log N x) 2 J „ lo yi v 7 J x v 7 -1 lnx 

A Un 8 / 

71. J^i dt= [In |t|]j nx = ln|lnx| -In 1 = In (lnx), x > 1 

72. f* \ dt = [In |t|] i = In e x - In 1 = x In e = x 

73. / L ' /X i dt= [ln|t|]I /x =ln |±| - In 1 = (In 1 - In |x|) - In 1 = -lnx, x >0 

74. J- f X i dt = [J- In Itll x = lnx ^ |nl = l g X, X > 

In a J i t Lin a I U l In a In a fea ' 

75. A = f_ 2 j^ dx = 2/ ^j dx; [u = 1+ x 2 => du = 2x dx; x = => u = 1, x = 2 => u = 5] 

-► A = 2 J' I du = 2 [In |u|] \ = 2(ln 5 - In 1) = 2 In 5 



76. A = J^ 2('- x) dx = 2 J^ (i) x dx = 2 



kD 



2 n 



-2-) C_i) = J_ 

ln2M 2^ ln2 



77. Let [H 3 + ] = x and solve the equations 7.37 = — log 1() x and 7.44 = — log| x. The solutions of these equations 
are 1(T 7 - 37 and 1CT 744 . Consequently, the bounds for [H 3 0+] are [1(T 744 , 1CT 7 - 37 ] . 

78. pH = - log K) (4.8 x 1(T 8 ) = - (log 10 4.8) + 8 = 7.32 

79. Let O = original sound level = 10 log 10 (I x 10 12 ) db from Equation (6) in the text. Solving 

O + 10 = 10 log 10 (kl x 10 12 ) fork => 10 log 10 (I x 10 12 ) + 10 = 10 log 10 (kl x 10 12 ) => log 10 (I x 10 12 ) + 1 
= log, (kl x 10 12 ) => log 10 (I x 10 12 ) + 1 = log 10 k + log 1() (I x 10 12 ) => 1 = log,„ k => 1 = ££ 
=> In k = In 10 =4> k = 10 

80. Sound level with 101 = 10 log 10 (101 x 10 12 ) = 10 [log 10 10 + log 10 (I x 10 12 )] = 10+10 log 10 (I x 10 12 ) 

= original sound level + 10 => an increase of 10 db 



81. (a) If x = [H 3 0+] and S - x = [OH"] , then x(S - x) = 10~ 14 => S = x + ^ => jjf = 1 - ^ 

d 2 S _ 2-10- 14 



and 



d\- 



> => a minimum exists at x = 10 



(b) pH=-log 10 (10- 7 )=7 



(c\ [° H ~] - S^x 
W r H ,0+] - x 



(x+^H)-x 



10" 



the ratio L Q l equals 1 at x = 10 



-7 



82. Yes, it's true for all positive values of a and b: log a b = p* 2 and log b a = pr =^> p^- = ^ = log., b 



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450 Chapter 7 Transcendental Functions 



83. From zooming in on the graph at the right, we estimate 
the third root to be x w —0.76666 




X-- 0.76666 



84. The functions f(x) = x ln2 and g(x) = 2 hlx appear to 
have identical graphs for x > 0. This is no accident, 



because x In2 = e lr 



l n2 \lnx 



(e ln2 ) 




85. (a) f(x) = 2 X ^> f'(x) = 2 X In 2; L(x) = (2° In 2) x + 2° = x In 2 + 1 « 0.69x + 1 
(b) 




/ y = (ln2)i + l 



-3 7T—A li 1 i 1 





1.4 


// 




1.2 


// 

/ j/ = (ln2)x + l 


y = 2* 

•7 


/0.8 






0.6 
0.4 
0.2 




\ -0.5 




o ■ ■ 0:3 ■ ■ -I x 



86. (a) f(x) = log 3 x => f '(x) = ^ , and f(3) = £§ => L(x) = 3^ (x- 3) +£§ = 3^-^ + 1 
w 0.30x + 0.09 
(b) 




87. (a) log 3 8 



In 8 
to 3 



1.89279 



(c) log 2() 17 = %% » 0.94575 

(e) In x = (log 1() x)(ln 10) = 2.3 In 10 « 5.29595 

(g) In x = (log 2 x)(ln 2) = -1.5 In 2 « -1.03972 



88. (a) ^•log 1 „x=>^. 1 ^ ) = ^=log 2 x 



> 

1.2 




^(x-Si-M^ 


1.1 




y/ y = log 3 x 


1 






0.9 






0.8 


//' 




0.7 


// 






/ 




2 2:5 


3 3:5 4 * 




(b) log 7 0.5 = ^ 






(d) log„, 7 = £fc 



-0.35621 

-2.80735 

(f) In x = (log 2 x)(ln 2) = 1.4 In 2 w 0.97041 
(h) In x = (log 10 x)(ln 10) = -0.7 In 10 « -1.61181 



In a ]„„ „ _ ln_a lnjc _ ln_x _ 1 

In b In a In b lu & b A 



(b) ^-10g a X: 



1-2 



dx V 2' 



k) = -xandA(i nx + c ) = 1. 



Since — x ■ - = — 1 for any x^0, these two curves will have perpendicular tangent lines. 
90. e ln x = x for x > and ln(e x ) = x for all x 

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Section 7.5 Exponential Growth and Decay 451 



91. Using Newton's Method: f(x) = ln(x) - 1 => f (x) = 1 =4> x n+1 = x n - ln( *f 1 =>■ x n+1 = x n [2 - ln(x n ; 

Then, Xi = 2, x 2 = 2.61370564, x 3 = 2.71624393, and x 5 = 2.71828183. Many other methods may be used. For example, 
graph y = In x — 1 and determine the zero of y. 

92. (a) The point of tangency is (p, In p) and m tange nt = - since £ = j. The tangent line passes through (0, 0) =>• the 

equation of the tangent line is y = -x. The tangent line also passes through(p, In p) => In p = -p = 1 =>■ p = e, and 
the tangent line equation is y = ^x. 

(b) jp = — \ for x^0=^y = lnxis concave downward over its domain. Therefore, y = In x lies below the graph of 

y = -x for all x > 0, x ^ e, and In x < - for x > 0, x ^ e. 

j e ' / e / 

(c) Multiplying by e, e In x < x or In x e < x. 

(d) Exponentiating both sides of In x e < x, we have e' nxe < e x , or x e < e x for all positive x/e. 

(e) Let x = 7r to see that if < e n . Therefore, e 51 is bigger. 

7.5 EXPONENTIAL GROWTH AND DECAY 

1. (a) y = y e kl => 0.99y = y e loook => k = l -^ « -0.00001 

(b) 0.9 = e ( - 000001)1 => (-0.00001)t = ln(0.9) => t = _'g^ w 10,536 years 

(c) y = y e( 20 ' 00 °) k w y e- (U = y (0.82) => 82% 

2. (a) % = kp => p = p e kh where p = 1013; 90 = 1013e 2Ok =>■ k = '"(*>) -to (ioi3) a -0.121 

(b) p = 1013e- 6 - 05 « 2.389 millibars 

(c) 900 = 1013e(-°- 12 ') h =>. -0.121h = In ($£) => h = ln ffl j ^ w 0.977 km 

3. I = -0.6y => y = yoe" - 61 ; y = 100 => y = lOOe" - 61 =>• y = lOOe" 06 w 54.88 grams when t = 1 hr 

4. A = A e kl =^> 800 = 1000e lok => k = "^ =>• A = 1000e (ln(,, - 8 >'' 10 >, where A represents the amount of 
sugar that remains after time t. Thus after another 14 hrs, A = 1000e (ln(0 - 8)/10)24 w 585.35 kg 

5. L(x) = L e- kx =>• ^ = L e- 18k =4> In | = -18k => k=^« 0.0385 =4> L(x) = L e- 00385x ; when the intensity 
is one-tenth of the surface value, ^ = L e-° 0385x => In 10 = 0.0385x => x « 59.8 ft 

6. V(t) = Voe"'/ 40 =4> 0.1V =V e- 1 / 40 when the voltage is 10% of its original value =>■ t= -40 In (0.1) 

w 92.1 sec 

7. y = y e kl and y = 1 =>• y = e kl => at y = 2 and t = 0.5 we have 2 = e°- 5k => In 2 = 0.5k =>■ k = £f = In 4. 
Therefore, y = e( ,n4 > =>■ y = e 241n4 = 4 24 = 2.81474978 x 10 14 at the end of 24 hrs 

8. y = y e kl and y(3) = 10,000 => 10,000 = y e 3k ; also y(5) = 40,000 = y e 5k . Therefore y e 5k = 4y e 3k 

=>. e 5k = 4e 3k => e 2k = 4 =4> k = In 2. Thus, y = yoe('° 2 »' =4> 10,000 = y e 31n2 = y e ,n8 =4> 10,000 = 8y 
^ yo = io|00 = 1250 

9. (a) 10,000e k « = 7500 => e k = 0.75 => k = In 0.75 and y = 10,000e< In,, - 75 \ Now 1000 = 10,000e( lnO - 75 >' 



In 0.1 = (In 0.75)t => t = jjj^L ~ g Q0 years (to the nearest hundredth of a year) 
1 
year) 



In 0.75 

, n T^'it -v. t — _ 

In 0.75 



(b) 1 = 10,000e (ln0 - 75 > => In 0.0001 = (In 0.75)t => t = lj \W^ ~ 32.02 years (to the nearest hundredth of a 



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452 Chapter 7 Transcendental Functions 

10. (a) There are (60)(60)(24)(365) = 31,536,000 seconds in a year. Thus, assuming exponential growth, 

P = 257,313,431e kl and 257,313,432 = 257,313,431e( 14k / 31 ' 5,6 ' 00 °) => In ( l|fj|ff ) = M^fooo 

=4> k« 0.0087542 
(b) P = 257,313,431e (00087542) ( 15 ' ss 293,420,847 (to the nearest integer). Answers will vary considerably 
with the number of decimal places retained. 

11. 0.9Po = Poe k => k = In 0.9; when the well's output falls to one-fifth of its present value P = 0.2P 

=> 0.2P = P e('° M > => 0.2 = e( lnM > => In (0.2) = (In 0.9)t ^ t = |§f « 15.28 yr 



12- (a) d i 



100 P 



r(x) 
2000 



f = - yig dx => lnp = - T i 5 x + C =► p = e(-°- 0l *+V = e c e -o.oix = Cie -o.«»ix. 

p(100) = 20.09 => 20.09 = Cie ( -°- 01)(100) => Cj = 20.09e rs 54.61 => p(x) = 54.61e-° 0,! ' (in dollars) 

(b) p(10) = 54.61e(-°- 01 »( 10 ) = $49.41, andp(90) = 54.61e ( - ,u,1 '( 90 ) = $22.20 

(c) r(x) = xp(x) => r'(x) = p(x) + xp'(x); 
p'(x) = -.5461e- ,u,lx => r'(x) 

= (54.61 - .5461x)e-°- 01s . Thus, r'(x) = 

=> 54.61 = .5461x => x = 100. Since r' > J6CK 

for any x < 100 and r' < for x > 100, then 
r(x) must be a maximum at x = 100. 




50 100 150 200 



13. (a) A e(°- 04 ) 5 = A e - 2 

(b) 2A = A e (0 - 04 > => In 2 = (0.04)t 
=> t = -£1 w 27.47 years 



In 2 
0.04 



17.33 years; 3A = A e (0 - 04)l ^ In 3 = (0.04)t 



14. (a) The amount of money invested A after t years is A(t) = A e' 

(b) If A(t) = 3A , then 3A = A e' => In 3 = t or t « 1.099 years 

(c) At the beginning of a year the account balance is Aoe', while at the end of the year the balance is Aoe (t+1) . 
The amount earned is Aoe (1+1 ' — Aoe 1 = Aoe'(e - 1) « 1,7 times the beginning amount. 

15. A(100) = 90,000 => 90,000 = lOOOe'C 00 ' => 90 = e l0,,r => In 90 = lOOr => r = ^ ~ 0.0450 or 4.50% 



16. A(100) = 131,000 => 131,000 = 1000e 100r => In 131 = lOOr => r 

17. y = yoe -0181 represents the decay equation; solving (0.9)yo = yoe -0 - 181 



^ w 0.04875 or 4.875% 



=> t 



In (0.9) 
-0.18 



0.585 days 



18. A = A e kl and i A = A e 139k => \ = e lwk => k = "^ « -0.00499; then 0.05 A = Aoe- - 0049 " 



139 



In 0.05 
-0.00499 



600 days 



19. y = y e" 



yoe- 



-(k)(3/k) 



Yoe" 



-3 _ yo ^ yo 



< In = (0-05)(yo) => after three mean lifetimes less than 5% remains 



e 3 ^ 20 



20. (a) A = A e- k ' => \ = e" 2 - 645k ^ k = ^ « 0.262 
(b) 1 w 3.816 years 



(c) (0.05)A = Aexpi 



In 2 
2.645 



t) => _i n20 =(-^)t => t 



2.645 In 20 
In 2 



11.431 years 



21. T - T s = (T - T s )e- k ', T = 90°C, T s = 20°C, T = 60°C => 60 - 20 = 70e-' 0k => | = e" 



10 



0.05596 



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Section 7.6 Relative Rates of Growth 453 

(a) 35 - 20 = 70e" ' 0559st => t w 27.5 min is the total time =>■ it will take 27.5 - 10 = 17.5 minutes longer to reach 

35°C 

(b) T - T s = (T - TJe" 1 ", T = 90°C, T s = -15°C => 35 + 15 = 105e- W)559 <" => t « 13.26 min 

22. T - 65° = (T - 65°)e- kl => 35° - 65° = (T - 65°) e" 10 " and 50° - 65° = (T - 65°)e- 20k . Solving 

-30° = (T - 65°)e- 10k and -15° = (T - 65°)e- 20k simultaneously => (T - 65°) e" 101 ' = 2(T - 65°)e- 20k 
=> e 10k = 2 => k = ^ and -30° = ^^ => -30° [e 10 ^) ] = T - 65° => T = 65° - 30° (e 1 " 2 ) = 65° - 60° = 5° 



23. T - T s = (T - T s ) e" 1 " => 39 - T s = (46 - T s ) e" 10k and 33 - T s = (46 - T s ) e" 20k => 



20k _^ 39-T, 
46-T, 



e" 10k and 



/ p -i0k\2 33-T, _ / 39-T, 

l c ) =" 46-T ~~ \ 46-T, 



33-T, _ 20k 
46-T s ~~ c 

= 1521 - 78T S + T 2 => -T s = 3 => T s = -3°C 



(33 - T s )(46 - T s ) = (39 - T s ) 2 =>■ 1518 - 79T S + T 2 



24. Let x represent how far above room temperature the silver will be 15 min from now, y how far above room 
temperature the silver will be 120 min from now, and to the time the silver will be 10°C above room 
temperature. We then have the following time-temperature table: 



time in min. 





20 (Now) 


35 


140 


to 


temperature 


T s + 70° 


T s + 60° 


T s + x 


T s + y 


T s + 10° 



T - T s = (T - T s ) e- kl => (60 + T s ) - T s = [(70 + T s ) - T„] e" 20k => 60 = 70e- 20k => k = (- ^) In (f ) 
w 0.00771 

(a) T - T s = (T - T s )e- ,U)077U =^> (T, +x) -T, = [(70 + T.) - T,] e - ((u,0771 >( 35 ) =^ x = 70e" 

(b) T - T s = (T - TJe-"- 007711 => (T, + y) - T s = [(70 + T.) - Tj e -( - ,,0771 )( 14 ") ^ y = 70e 

(c) T - T s = (T - T s )e- ,U)077U => (T, + 10) - T, = [(70 + TJ - T,] e -<°- 00771 ) l ° =>• 10 = 70e- ,u,077Uil 



o.269 85 _ 53.440c 
-1.0794 r^, jo 7Q°r 1 



In 



-0.00771t => t 



7 y ^.w«,, iH , —r hj y 0.00771 

silver will be 10°C above room temperature 



In 



252.39 => 252.39 - 20 « 232 minutes from now the 



25. From Example 5, the half-life of carbon-14 is 5700 yr 



c = coe 



(0.445)c = c e- 



i c = c e- k ( 57 ™) 

In (0.445) 
-0.0001216 



k= ±± re 0.0001216 



6659 years 



26. From Exercise 25, k re 0.0001216 for carbon-14. 

(a) c = coe-"- 00012161 => (0.17)c = coe-"- 00012161 => t re 14,571.44 years 

(b) (0.18)c = coe- - 00012161 => t re 14,101.41 years => 12,101 BC 

(c) (0.16)c =c e- 000012 "" =>• t re 15,069.98 years =>■ 13,070 BC 



12,571 BC 



27. From Exercise 25, k re 0.0001216 for carbon-14. Thus, c = Coe 



In (0.995) 
-0.0001216 



41 years old 



(0.995)c = c e- 



7.6 RELATIVE RATES OF GROWTH 



1. (a) slower, lim ^^ — \i m J_ = q 

x ^ oo e x ^ oo e 

(b) slower, lim * 3 + f 2 * = lim 3 " 2 + 2 g x cos x = lim 6x + 2 x cos2x = lim 6 ~ 4 f n2x = by the 

x ^ oo e x ^ oo e x ^ oo e x ^ oo e J 

Sandwich Theorem because 4 < 6 ~ 4 ' in2x < ^ for all reals and lim 4=0= lim *Q 



(c) slower, lim -v = lim ^ = lim 

X ^ OO e X — > OO e x ^ oo 



(d) faster, lim § = lim 

X — > CO e X — > oo 



e x 
f4\ x 



§ s- 1/2 



1 



lim , - 

x — > oo V xe 







oo since - > 1 



(e) slower, lim ^f- = lim (^ 

v y X — » OO e X — > oo v 2e 



) x = o 



since ^ < 1 



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454 Chapter 7 Transcendental Functions 



(f) slower, lim 

X — > oo 



lim 



OO e x . 



(h 

2. (a 
(b 

(c 

(d 

(c 
(f) 
(g 

(h 

3. (a 
(b 

(c 

(d 

(c 
(f) 
( 
(h 

4. (a 
(b 
(c 

(d 

(c 

(f) 

( 

(h 



same, lim 

X — > oo 

slower, lim 

X — » oo 



slower, lim 

x — > oo 



slower, lim 

X — » oo 



= lim ^-r- 

X — > oo e 

slower, lim 

X — > oo 

= ^0 = 



slower, lim 

X — > oo 

slower, lim 

X — » oo 

faster, lim - 

x — > oo 



lim k 

X — » oo 2 



logio x 



lim 7r % x - 
X — > oo ( ln 10 ) e 



X — > OO On 10 ) e 



lim ,, ,L x 

X — ► OO ( ln 10)xe x 



10x 4 + 30x+l 



lim 

X — » oo 



40x 3 + 30 



lim 

X — » oo 



xmj^x = Um 

e X — » oo 

= lim -^ = 

X — » OO xe" 
\/l+x 4 _ 



x(lnx- 1) 



lim 

X — > oo 



120x 2 

e x 



lim 

X — > 00 



2 -Mix 



lim 

X — > 00 



lim 

X — v oo 

ln x - 1 + 1 



240 



lim 

X — > oo 



lim 4# 

X — > oo e 



lim M 

X — ► oo 2e 



lim , 

X — > oo 4e 



12x2 



lim |% 

X — > OO 8e 2x 



24 



lim 

X — > 00 16e 



= lim (f 

x — > oo V2e ^ 

= lim 4r = 

X — > oo e 

lim x = oc 

X — > oo 



since £ < 1 



slower, since for all reals we have — 1 < cos x < 1 



< e c ' 



< e 1 => 



< 



lim <= 

X — ► OO ' 

same, lim 

X — » oo 







lim 

x-»oo 



so by the Sandwich Theorem we conclude that lim 

J X — > oo 



< % and also 

5^ = 



lim 

X — > OO e ' 



(x-x+1) 



lim i 

X — > oo e 



same, lim 

X — ► oo 

faster, lim 

X — > oo 

same, lim 

X — > oo 

same, lim 

X — » oo 



x^ + 4x 



lim ^ti 

X — > OO 2x 

: lim (x 3 - 1) 

X — > oo v 7 



lim 

X — > oo 



\/x 4 + > 



(x + 3) 2 



lim 

X — > oo 



lim 

X — > oo 



:i + x) = v^i 



1 = 1 



lim 

X — > oo 



2(x + 3) 



slower, lim 

X — > oo 

faster, lim \ 

X — > OO x 2 

slower, lim — 



same, lim %- 

X — > OO x 2 



same, lim 

X — > oo 



x ln x 

y2 



lim 

X — > oo 



lim 

X — > oo 



2x 

ln x 
x 

(ln 2) 2 X _ 



lim i 

X — » oo 2 



2x 



: = lim 4 : 

x — ► oo e 

lim 8 = 8 

X — > 00 



lim 

X — > oo 

lim 

X — > oo 

lim 



l 

(In 2) 2 2" 



same, lim 

X — » oo 

slower, lim 

X — > oo 



slower, lim 

X — > oo 



10x 2 



= lim ( 

X — ► 00 v 

lim 10 = 10 

x — > oo 

= lim i = 

X — > oo e 



x3/2/ 



log 10 x 2 

X 2 
3_ v 2 



lim 

X — > oo 



. In 10 I 



f^ta lim 

In 10 x — > OO 



2 In x 
x2 



r4n lim 

In 10 x — > 00 



(i) 

2x 



rAn lim 

In 10 x — ► OO 



faster, lim 

x — > oo 



slower, lim 

X — > oo 



faster, lim ^^- = 

X — > OO x 2 

same, lim x + j 00x 

X — > OO x 2 



lim (x — 1) = oo 

X — » oo 




= x^oc «=° 




lim (•"'•'XiD x = n m 0" 

X — > OO 2x X — ► 00 


.D 2 (i.i) x 

2 


= lim (l + 1Q0) = i 

x — y oo v x 1 





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Section 7.6 Relative Rates of Growth 455 



5. (a) same, lim ^ 



lim 

X — > oo 



(Ml 

In x 



lim 

X — > oo 



1 

In 3 



1 
In 3 



(b) same, lim ! j-^ = lim 

W X ^ OO ln x x ^ oo 



(£) 



(c) same, lim -p^ 

v ' X — > OO In x 



(d) faster, lim 

X — > oo 

(e) faster, lim 

X — > oo 



In. 

lim ^ 

X — > oo ln x 



lim k = k 

X — > OO 2 2 



lim f = lim ffl£- 

X — > OO lnx X — > OO (1) 



lim tt^ = lim 

X — > OO 2yx x — » OO 



^ 



CO 



lim 

X — > oo 



1 



lim x = oo 

X — > oo 



(f) same, lim ^^ = lim 5 = 5 

w x ^ oo ln x x ^ oo 



(e) slower, lim -r^- = lim — f— 

V&/ ' X — > oo ln x x — > oo x ln x 



(h) faster, lim 



lim ff 

X — > OO (; 



lim xe x 

X — > oo 



6. (a) same, lim ^^ = 

V ' X — > OO ln x X 



lim 



Mix 2 '. 

v'" 2 ) 

ln x 



1 i: „ ln x 2 

t—f; lim -j 

ln 2 x — > OO ln x 



r^ lim ^ 

ln 2 X — > OO ln x 



(b) same, lim 

X — » OO 



logio 10* 
ln x 

1 



n ]l)x *) 
In 10 I 



lim , 

X — > OO ln x 



-X_ ii m lnl0x 

In 10 x — » OO ln x 



In 10 x — ► OO 



(c) slower, lim / = 



lim 



OO (\A)(bix) 



r^ lim 2 

ln 2 x — > OO 



2 
In 2 



,-Vn lim 1 = r^rs 

In 10 x — ► OO ™ ^ 



(d) slower, lim 4=^- 

V ' X — > OO ln x 



lim 



OO x2 In x 



(e) faster, lim 

x — > oo 



x— 2 ln x 
ln x 



lim 

X-tOO 



(r L --2) = ( lim r~) ~ 2 

Unx I \ x _> oo lnxy 



lim 

x — > OO I I N 



-2 



( lim x ) 

\x — ► oo / 



(f) slower, lim £— 

V ' X — » oo ln x 



lim -^— 

X — > OO e ln x 



(g) slower, lim ^2^ 

V&/ X — > oo ln x 



lim 

X — > oo 



u*) 



lim ri- 

X — ► OO ln x 



(h) same, lim n , + 

v ' ' X — > OO In x 



lim 

X — > oo 



lim 



2x 

bo 2x + 5 



lim 



OO 2 



lim 1 = 1 

X — > oo 



7. lim -§, 

X — > OO e ' 

= lim 

X — > oo 



lim e x / 2 = oo 
X — > oo 



e x grows faster than e x / 2 ; since for x > e e we have ln x > e and lim 

e X — > oo 



(ln x) x 



' ln x \ x 



(In x) x grows faster than e x ; since x > ln x for all x > and lim 



lim 

X — > oo 



oo => x x grows faster than (ln x) x . Therefore, slowest to fastest are: e x / 2 , e x , (ln x) x , x x so the order is d, a, c, b 



lim 

X — > oo 



(In 2)" 



lim 

X — > oo 



(ln (ln 2))(ln 2)" 
2x 



(In 2) x grows slower than x 2 ; lim 



lim 

x — ► oo 

V 2 



(In(ln2)) 2 (ln2) x _ (ln(ln2)) 2 



2 



lim (ln 2) x = 

x — > oo 



2x 



lim ..,,,. 

X — > OO ( ln 2)2 X 



X^OO 0n2) 2 2 x 



=> x 2 grows slower than 2 X ; 



lim 

X — > oo 



lim 

X — > oo 



x — > oo 
e) = =^ 2 X grows slower than e x . Therefore, the slowest to the fastest is: (ln2) x ,x 2 ,2 x 



and e x so the order is c, b, a, d 



(a) false; lim - = 

v ' X — > oo x 

(b) false; lim -^ 

w X — > oo x + 5 

(c) true; x < x + 5 

(d) true; x < 2x => 



2x 



j^= < 1 if x > 1 (or sufficiently large) 
< 1 if x > 1 (or sufficiently large) 



(e) true; lim -% 

X — » oo e 



(f) true: 



x + In x 



X 
ln x 



lim 
< 1 







\-\ — 7= < 2 if x > 1 (or sufficiently large) 



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456 Chapter 7 Transcendental Functions 



(g) false; lim fe = Inn 



ln2x 



,^4- = lim 1 = 1 



(h) true; ^p^ < ^ {x + 5)2 < >i±5 = i + | < 6 if x > 1 (or sufficiently large) 



10. (a) true; /{\ = ^3-3 < 1 if x > 1 (or sufficiently large) 
w 

(b) true; 44 = 1 + j < 2 if x > 1 (or sufficiently large) 



(c) false; lim 



v> i 2 J 



x -» 00 (l) 

(d) true; 2 + cos x < 3 - : °" x 3 

(e) true; ^ 



lim (1 - i) = 1 

x — > 00 v xy 



< ^ if x is sufficiently large 
1 + 4 and 4 — > Oasx — > 00 =>• l + 4<2ifxis sufficiently large 



(f) true; lim ^ = lim ^ = lim -^ 

X ^ OO x x^oo x x — > 00 1 

(g) true; -221) < |f = 1 if x is sufficiently large 







(h) false; lim 



lim 



X — > OO ln(x 2 +l) x _> co (2* 



lim 4^ 

X — > cx> 2x2 



lim (i + =ij) = ± 

X — > (X) v 2 2x y 2 



11. If f(x) and g(x) grow at the same rate, then lim 44=1-7^0 => lim 

v ' 6V y 6 x — > 00 sW ~ x — > c 



M _ T a. n - 1™ SW - . I ^ 0. Then 



OO « x > L 



L < 1 if x is sufficiently large =>L-1<-^<L+1 => ^ < |L| + 1 if x is sufficiently large 



|g(X) 
=>• f = 0(g). Similarly, jg < |±| + 1 => g = 0(f). 



fix) 



12. When the degree of f is less than the degree of g since in that case lim 44 = 0. 

b b b x _> qq g( x ) 



fix) 



13. When the degree of f is less than or equal to the degree of g since lim 44 = when the degree of f is smaller 

b 1 fc & x _> oq g(x) b 



f(x) 



than the degree of g, and lim 44 = § (the ratio of the leading coefficients) when the degrees are the same. 



14. Polynomials of a greater degree grow at a greater rate than polynomials of a lesser degree. Polynomials of the 
same degree grow at the same rate. 



15. lim 

x — > 00 



ln(x+l) 
In x 



x+1 , 



lim 

x — >■ 00 (i 



lim -iv = lim i = 1 and lim '"< x + " 9) 
x— > 00 x+1 x^oo 1 x— > 00 lnx 



lim 

x — > 00 



X + 999 J 



lim , x „ nn 

X — » OO x + 999 



16. lim lr 4°^ 
x — > 00 i |lx 



lim 

X — > CO 



. x + a 



lim -f- 

X — > CO x + a 



lim 1 

X-tOO ' 



1. Therefore, the relative rates are the same. 



17. lim 

x — > 00 



4iQx+i 



lim 

X — > CO 



10x+l 



10 and lim 

x — » 00 



/x+l 



lim =ti = v 1 = 1. Since the growth rate 

x — > 00 x v 6 



is transitive, we conclude that x/lOx + 1 and \/x + 1 have the same growth rate (that of x/x) 

4 x4 + x _ 



18. lim 

x — *• 00 x ' 



lim *-+* = 1 a nd lim ^f^ 

X — > OO x X — » 00 x 



lim - / 

x — > 00 x 



1 . Since the growth rate is 



transitive, we conclude that \J x 4 + x and y x 4 — x 3 have the same growth rate (that of x 2 ) . 



19. lim 4 

x — > 00 e 



lim 

x — > 00 



lim 4=0 

X — > CO e 



x n = o (e x ) for any non-negative integer n 



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Section 7.6 Relative Rates of Growth 457 



20. Ifp(x) = a n x n + a n _ix n 



aix + an, then lim 

u X — > oo 



p(x] 



a n lim ^r + a n _ i lim 

x — > OO e X — *• oo 



p(x) 



ai lim 4 + an lim 4 where each limit is zero (from Exercise 19). Therefore, lim - 1 -— = 
x ^ co e u x ^ co e x ^ co 

$■ e x grows faster than any polynomial. 



21. (a) lim ^ = lim s^ 

v/ X^OO lnx X — > OO n(i) 



i) lim x 1/n 

n ' X — > CO 
1/10 6 



(b) m (e 17 > 000 < 000 ) = 17,000,000 < (e 17xl ° 6 ) 

(c) x« 3.430631121 x 10 15 

(d) In the interval [3.41 x 10 15 ,3.45 x 10 15 ] we have 
In x = 10 In (In x). The graphs cross at about 
3.4306311 x 10 15 . 



IT 



co =4> In x = o (x 1 ' n ) for any positive integer n 
i 24,154,952.75 



0.004' 


y = In x- 10 ln(ln x) 




0.002 






-0.002 


3.41 io y^ 


3.45 10 


-0.004 


yS 




-0.006 


/ 





22. lim 

X — > CO 

lim 



In x 



lim af 

noo V x ) 



a„x" + a„_, x n_l + . . . + ajx + ag 



(■!) 



lim 



x — > oo M ( nx ") 



=> In x grows slower than any non-constant polynomial (n > 1) 



23. (a) lim 



n log2 n 
CO n (logi n) 2 



lim 



OO lo E2 n 



=> n log2 n grow$>) 



slower than n (log? n) ; lim 

° n — > oo 



n log2 n 

n3/2 



I — ] 

hm -^7/- 
n — > co n ' 



r^TT lim 
In 2 n 



, W _ _2_ lim J_ 

CO (i) n-V2 - In 2 n ™CO "1/2 







=4> n log2 n grows slower than n 3,/2 . Therefore, n log2 n 
grows at the slowest rate => the algorithm that takes 
0(n log2 n) steps is the most efficient in the long run. 




24. (a) lim 

n — > co 



(log, n) 2 



lim 

n — > co 



lim ^% 

n — > oo n < ln 2 r 



,. 2 < lnn )(n) 

lim ,. .A 7 

n — » oo ( ln 2 ) 2 

lim 



lim 



(In 2)2 n^+ oo i 



(ln 2) 2 n _> go n 



=>• (log2 n) 2 grows slower 



than n; lim ^2ip>L = lim ^ 
n — > oo v n lo S2 n n — > oo v n 



I — 1 
i • V in 2 ; 

hm ^ 

n — > CO n 1/2 



r^> lim 

ln 2 n — > oo n 



In n 

1 2 



(b) 




20 40 60 80 100 



= r^ lim , n v " ,,, = r4; lim -k= = =>- (log2 n) grows slower than \/n log2 n. Therefore (log2 n) grows 

In 2 x — ► OO (2) n ' ln 2 n — > CO 1 ' v 

at the slowest rate => the algorithm that takes O ((log2 n) 2 ) steps is the most efficient in the long run. 

25. It could take one million steps for a sequential search, but at most 20 steps for a binary search because 
2 19 = 524,288 < 1,000,000 < 1,048,576 = 2 20 . 

26. It could take 450,000 steps for a sequential search, but at most 19 steps for a binary search because 
2 18 = 262,144 < 450,000 524,288 2 19 . 



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458 Chapter 7 Transcendental Functions 
7.7 INVERSE TRIGONOMETRIC FUNCTIONS 



1. (a) 



(b) 



(c) 



3- (a) -| 


(b) f 


(c) -f 


5. (a) f 


(b) 3 f 


(c) 1 


7. (a) 3 f 


(b) f 


(c) 2 f 



9. (a) I 



(b) -\ (c) | 



11. (a) 



3tt 



(b) I 



(c) 



2tt 



2. (a) 



4. (a) | 



6. (a) & 



8. (a) | 
10. (a) -f 
12. (a) 5 



(b) \ 



(b) 



(b) I 



(b) 



(b) | 



(b) 



5- 



(C) 



(c) I 



(0 f 



(c) f 
(c) f 



13. a = sin : (-^) =4> cos a = ||, tan a = ^, sec a = ||, esc a = y, and cot a = y 

14. a = tan -1 (|) =>■ sin a = I, cos a = 1, sec a = |, esc a = |, and cot a = | 



sin a = -75, cos a = \-, tan a = —2, esc a = ^-, and cot a 



sin a 



, cos a 



j=, tan a = — |, esc a = ^— , and cot a = — | 



18. sec (cos" 1 I) =sec(f) =2 



15. a = sec : 

16. a = sec" 1 (- ^) 

17. sin [cos- 1 &\ = sin (f) = 4- 

- 1 (-I))=tan(-I) = -^ 20. cot(sin- 1 (-f))=cot(-f) 

sec -1 2) + cos (tan- 1 ( — \J 3] j = esc (cos -1 (|)) + cos 

sec" 1 1) + sin (esc- 1 (-2)) = tan (cos" 1 \) + sin (sin" 1 (-5)) = tan(0) + sin (- |) =0+ (- |) = - | 



19. tan 



21. esc 



22. tan 

23. sin 

24. cot 



1 
v/3 



|)=csc(f) + cos(-f) = ^ + i = ^# 



sin- 1 (-I)+cos- 1 (-i))=sin(-| + f) 



sin(f) = l 



sin" 1 (- I) - sec" 1 2) = cot (- f - cos" 1 (±)) = cot (- f - 



i) =«*(-!) =0 



25. sec (tan -1 1 + esc -1 1) = sec (f + sin -1 \) = sec (f + |) = sec 

26. sec (cor 1 ^3 + esc" 1 (-1)] = sec (f + sin" 1 (i)) 

27. sec" 1 (sec (- |)) = sec" 1 (-4) - — - 1 ^ 

28. cor 1 (cot (- I)) = cor 1 (-1) = 



: 3 f) 



TT_ TT_ 7T1 

2 3 2) 



-V~2 



?2E 
4 



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Section 7.7 Inverse Trigonometric Functions 459 



29. a = tan l | indicates the diagram 




sec (tan" 1 |) = sec a = V^±4 



30. a = tan : 2x indicates the diagram 





2x 


V4x 2 + 1/ 


v^o 1 





sec (tan : 2x) = sec a = y 4x 2 + 1 



31. a = sec : 3y indicates the diagram 




y] 9 f-i => tan (sec : 3y) = tan a = y^y 2 - 1 



32. a = sec l I indicates the diagram 




tan (sec 1 i J = tan a = ^-j — 



-l y\ 



33. a = sin x x indicates the diagram 




» =>- cos (sin : x) = cos a = y 1 — x 2 



34. a = cos x indicates the diagram 



35. a = tan J v x 2 — 2x indicates the diagram 

V x 2 — 2x 




Vi-x 2 =>• tan (cos 1 x) = tan a 



\/l-x 2 



x' i -2x =^ sin 



(tan" 1 \/x 2 - 2x) 



sin a 



x- 1 



36. a = tan : , * indicates the diagram 




sin ( tan 1 , | ) = sin . 

\ \/ X 2 + 1 / 



\/2x 2 + 1 



37. a = sin : -y indicates the diagram 




=4> cos (sin : -j) = cos a = 3 y 



38. a = sin 1 | indicates the diagram 




=>• cos (sin : I) = cos a 



y/25^7 



39. a = sec x | indicates the diagram 




=>- sin sec 4 = sin a 



1 x\ - cln™ - \/x 2 -16 



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460 Chapter 7 Transcendental Functions 



40. a = sec 1 ^ x + indicates the diagram 





2 


Vx 2 + 4/ 


S^ix x 





■ ( -1 \/x 2 + 4\ 

sin I sec ^— ^ — J = sin . 



V^x 2 + 4 



41. lim sin x 

x^ 1- 



42. lim cos : x = 7r 

x->-l+ 



43. lim tan _1 x = g 

x — > oo 2 



44. lim tan -1 x 

X — > — oo 



45. lim sec x = 5 

X — > oo l 



46. lim sec : x= lim cos : (-) = % 

X — > — OO X — > — oo \^> * 



47. lim esc : x= lim sin : (-) =0 



48. lim esc J x= lim sin l (-) =0 

X — > — 00 X — » — 00 v "/ 



49. y = cos" 1 (x 2 ) => | 



2\ 



-2x 



/l - (x 2 ) 2 Vl^ 



50. y = cos l {-) = sec : x 



dy _ 1 

dx |x| Vx 2 -1 



51. y = sin-V2t => f - ------ 



1 - v/2t 



2 Vl-2t 2 



52. y = sin- 1 (1 - t) => S 



dy 



dt x/l-(l-t) 2 VSt^? 



53. y = sec" 1 (2s + 1) 



Jv 



|2s+l| v^s+l) 2 -! |2s+l| x/4s 2 + 4s |2s+l| x/s 2 + s 



54. y = sec 1 5s => 



dy 



|5s| x/(5s) 2 -l s| x/25s 2 - 1 



55. y = esc- 1 (x 2 + !)=>.£ 



2x 



-2x 



dx 



|x 2 +l| \/(x 2 +l) 2 -l (x 2 +l)v'x I T 



2\- 



dy 
dx 



56. y = esc : (|) 

57. y = sec -1 (1) = cos -1 1 



l!lv / (l? rT wV^ 4 ' |x|v/ *^ 



dy 
dt 



/ t _ t 2 



58. y = sin 



t 2 J 



CSC 



-(«) 



dy 
dt 



|.2| ./A? 



/t£-9 tvt 4 — 9 



59. y = cor 1 x/t = cor 1 1 1 / 2 => f 



iK^ 



l + (tV 2 ) 2 2x/t(! +0 



60. y = cor 1 x/t- 1 = cor 1 (t - l) 1 / 2 =*► g = - 



Q)<'-')- 1/2 = -i = -i 

l + [(t-l)!/ 2 ] 2 2^/1^1(1+1-1) 2tx/t--T 



61. y = In (tan l x) 



dy _ li+x^J 

dx tan -1 x 



(tan-lx)(l+x 2 



62. y = tan -1 (In x) 



dy = (j) = l 

dx l+(lnx) 2 x[l+(lnx) 2 ] 



63. y = esc- 1 (e') => g 



UA 



e'U/e'V-l 



x/e 2 ' - 1 



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Section 7.7 Inverse Trigonometric Functions 461 



64. y = cos x (e ') => 



dy 
dt 



v/l-(e-) 2 Vl-e" 2 ' 



65. y 



= svT^ 


S 2 


+ cos : s 

s 2 


= 8(1 


\/l-s 


1 


Vl-s 2 

- sec -1 s = 


l/l-S 2 


= Vs 2 - 

s |s|-l 


1 


= (s 2 - 



,1/2 



2\l/2 



.2N-1/2 



s (1 - s 2 ) v " + cos- 1 s => | = (1 - s 2 ) v/ + s (|) (1 - s 2 r v V2s) - 



V 7 ! - s 2 - 



s 2 + 1 _ 1 - s 2 - s 2 - 1 _ -2s 2 



'l-s 2 x/l-s 2 



M-s 2 



s| Vs 2 - 1 Vs 2 - 1 |s| Vs 2 - 1 



|s| Vs 2 -1 



67. y = tan- i yx 2 ^T + csc- 1 X = tan- 1 (x 2 -l) 1/2 + csc- 1 X =► g- iA 1 '^' ' "^ 



1+ ( 



x 2 -!) 1 ' 2 ] 2 Wv 7 ^ 3 ! 



x v x 2 — 1 |x| v x 2 — 1 



0, for x > 1 



68. y = cor 1 (!) - tan' 1 x = f - tarr 1 (x" 1 ) - tan" 1 x => ^ = =^U - -^ 

' U/ 2 v / dx l + (x _1 ) 1 + 



_ _J L_ - n 

-x 2 ~~ x 2 +l 1+x 2 — U 



69. y = x sin x x 



+ ^1-x 2 = x sin- 1 x + (1 - x 2 ) 1/2 => | = sin" 1 x + x (-7^) + (5) (1 - x 2 ) _1/2 (-2x) 



\7i-x 2 \/i~ 



70. y = ln(x 2 + 4)-xtan- 1 (|) => | = ^ - tan" 1 (f ) - x 



i+(ir 



2x 
x 2 + 4 



tan 



-1 (x\ _ 2x 

\2) 4 + x 2 



71. 



^-tan- 1 ^) 

/75b dx = sin_1 (f)+C 



72. f . x dx = i f . 2 , dx = i f^- , where u = 2x and du = 2 dx 

J V1-4X 2 2 J Vl-(2x) 2 2 J ^l- u 2 



! sin' 1 u + C = ! sin" 1 (2x) + C 



73- J TT ^dx=J 



(v/n) +x 2 



dx 



4- tan" 1 -£- + C 



74. /^ dx = |/p^ dx = ^ tan- 1 (^) + C = f tan" 1 (^) + C 



75. f , d * = f , d " , where u = 5x and du = 5 dx 

J xV25x 2 -2 J u\/u 2 -2 

= 73 sec "l73| +c= 75 sec "l^l +c 



76. f /* = f , d " , where u = J~5x and du = a/5 dx 

J x\/5x 2 -4 J uVu 2 -4 v v 

= isec- 1 |f|+C= isec- 1 |^P|+ C 
11 ■ II 707, = [ 4 sin_1 f] = 4 ( sin_1 I - sin_1 °) = 4 (I - °) 



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462 Chapter 7 Transcendental Functions 

,3\/2/4 



78 r 2/4 ^ = i r 

Jo V9 - 4s 2 2 Jo 



., , , du „ , where u = 2s and du = 2 ds; s = 

2 Jo \/9-u 2 

[I sin- |] f /2 = I (rin-'f - sin- o) = J (} - 0) = f 



n 3v/2 

u = 0, s = ~y 



3^2 



'2 dt; t = => u = 0, t = 2 ^> u = 2\/ 2 



79 - ThtW = jify&?> where u = ^ and du 

~ J - 1 (tan- 2j| - tan- o) = i (tan" 1 1 - tan" 1 0) = i (| - 0) = $ 



4s tan" 1 -jj- , 



80 - Il 2 4+W = 75 f-zJi 4+^P ' where u = V3t and du = v^ dt; t = -2 => u = -2a/3, t = 2 => u = 2^3 



81./. 



-V2/2 



dy 



-V5 

2 



p-\/2 , 

I — , " , where u = 2y and du = 2 dy ; y = — 1 =>• u= —2, y 

J -2 u\/u 2 ' 



[sec- 1 |u|] _a 2 = sec" 1 -a/2 - sec" 1 |-2| = f - f 



=> u= -V2 



-v^ 



82. f /n dy 2 ; = f /" , where u = 3y and du = 3 dy; y = - \ =» u = -2, y 

J -2/3 yx /9y 2 -l J -2 u\/u 2 -l ^ •"•' 3 > ■> 



-2/3 yV^y 2 - 1 J-i u\/u 2 -l 

[sec- 1 |u|]I 9 v/5 = sec" 1 l-y^l - sec" 1 |-2| 



4 3 



VI 



■y/2 



83. / 



3 dr 



,/l -4(r-l)2 2 J /T3 

= | sin -1 u + C = 2 



f , du , where u = 2(r - 1) and du = 2 dr 

J \/l-u 2 v ' 



3 sin- 1 u + C = | sin" 1 2(r - 1) + C 



84- / v ^ T7 =6/ v ^,whereu = r+landdu = dr 



6 sin -1 | + C = 6 sin" 



85. f , dx n2 = f ^S , where u = x - 1 and du = dx 

J 2 + (x- l) 2 J 2 + u 2 

= 72 tan "73 + c =73 tan "(7T) +c 

86 - / i + (3 d x+i) 2 = 3 JlTTP - where u = 3x + 1 and du = 3 dx 
= i tan" 1 u + C = i tan" 1 (3x + 1) + C 



87. / 



(2x-l) v /(2x-l) 2 -4 



i I — 4s — , where u = 2x — 1 and du = 2 dx 

1 J uvu 2 - 4 



! - ! — " l |u| ' C = isec- 1 1^1 



h 



J u\/u 2 



(x + 3)v'(x + 3) 2 -25 J u\/u 2 -25 

i _„-l Iu| , c= l sec -i v_: +c 



where u = x + 3 and du = dx 



I sec- 1 |^| 



89. r 2 rxr^il = 2 f ' -rrS . where u = sin 9 and du = cos 9 d0; 6> = - ? 

J _,r/2 1 + (sin 9) 2 J_i 1+u 2 ' ' 2 

= ptan-u]^ =2 (tan- 1 1 -tan^C-l)) = 2 [f - (- |)] = tt 



=^u= -1, 



|^u=l 



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Section 7.7 Inverse Trigonometric Functions 463 



90 - JT/6 Tf^ = -/^TT^' whereu = cotxanddu =- csc2xdx ; x =I = > u= v / 3,x=| :> u=l 
= [- tan -1 u] ] ^5 = - tan -1 1 + tan -1 \pb ■ 



7T _i_ 7T _7T_ 

4 ' 3 ~~ 12 



91. I 

Jo 



e* dx 



- | T^-o , where u = e x and du = e x dx; x = => u=l,x = ln v3 =^ u = v3 

1 + e?" J i 1 +U- 2 ' ' ' v v 

= [tan- 1 u] /* = tan" 1 a/3 - tan" 1 1 = § - * = jl 



92. f , n lf 2 ^ = 4 r t^ , where u = In t and du = ± dt; t = 1 => u = 0, t = e^/ 4 => u = f 

J I t ( 1 + m z t) J o 1 + u t 4 

= [4 tan" 1 u] q /4 = 4 (tan" 1 | - tan" 1 0) = 4 tan" 1 f 

93. f ^hr = 5 f^— . where u = y 2 and du = 2y dy 

•J \/i - y 4 2j v /|_ u 2 

= | sin -1 u + C = \ sin -1 y 2 + C 



94. / 



/ , du , where u = tan y and du = sec 2 y dy 



sec 2 y dy 
v/l-tan 2 y ~~ J /T 

= sin -1 u + C = sin -1 (tan y) + C 



J \/-x 2 + 4x-3 J 



dx 



./-x 2 + 4x-3 J N /l-(x 2 -4x + 4) J x /l-(x-2)2 



J: 



dx 



sin -1 (x - 2) + C 



96. f ^2= = f 

J \/2x-x 2 J 



dx 



x /l-(x 2 -2x+l) J Vl-(x-l)- 



J- 



dx 



sin" 1 (x - 1) + C 



97 - £ ya=? = 6 I V (*+»+.> = 6 L 7 w J ww = 6 [-*- 1 m] -i 



V3-21-1 2 J-i v / 4 -( t2 + 2t + 1 ) "J-ix/2 2 " 

6 [sin" 1 (i) - sin" 1 0] = 6 (| - 0) = tt 



I, 



' 6dt _ -x f ' 2dt 

Jl/ 



/2 ^/3 + 4t-4t 2 Ji/2 % /4-(4t 2 -4t+l) J 1/2 % /2 2 -(2t-l) 2 



3 r 1 - 

J 1/2 • 



2dt 



3 [-- 1 (3^)] I 



3 [sin 



l (i\ 



^0] =3(f-0) = f 



/2 



on P dy _ P dy _ P 

"• J y 2 -2y + 5 ~ J 4 + y 2 -2y+l — J 1 



, , d y _ I tan -i (y^A\ i r 

y 2 -2y + 5 ~~ J 4 + y 2 -2y+l — J 2 2 + (y-l) 2 ~~ 2 Um I 2 J ^ ^ 



loo- /.^no = / 



dy 



y 2 + 6y+10 J l + (y 2 + 6y + 9) J l + (y + 3) 2 



J, 



dy 



tan" 1 (y + 3) + C 



ioi. r^fn = 8 /, 2 



x 2 -2x + 2 - U J, !+(x 2 -2x+l) - 8 J, l+(x-l) 2 - 8 t tan 1( - X ~ ! )ll 

8 (tan- 1 1 - tan" 1 0) = 8 (f - 0) = 2tt 



i° 2 - r^^o= 2 r iK -' ,i iK 



€ 



2 x 2 -6x+10 J 2 l + (x 2 -6x + 9) J 2 l+(x-3) 2 

= 2 [tan- 1 1 - tan- 1 (-1)] = 2 [J - (- f )] = tt 



2 [tan- 1 (x- 3)] ^ 



103 f ^ = f ^ r 

J (x+l)Vx 2 + 2x J (x+l)Vx 2 + 2x+l-l J ( 



;+l)Vx 2 + 2x J (x + 1) Vx 2 + 2x + 1 - 1 J (x+ l^x + l) 2 - 1 

f — , d " , where u = x + 1 and du = dx 

J UA/U 2 — 1 

sec -1 lul +C = sec -1 |x + 1| + C 



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464 Chapter 7 Transcendental Functions 



dx I dx 

J (x- 



— =f- 

i + 4 - 1 J (x - 



J (x-2)\/x 2 -4x + 3 J (x-2)Vx 2 -4x + 4-l J (x-2)\/(x- 2) 2 - 1 

= / — ,\ du, where u = x — 2 and du = dx 

J uy u 2 — 1 



sec" 1 |u| + C = sec" 1 |x - 21 + C 



105. I e r" dx = I e" du, where u = sin 1 x 



and du 



\/l-x 2 



e u + C = e sin X + C 



106. | e , c °" dx = — e" du, where u = cos l x and du 



-dx 



yr 



-e" + C = -e cos x + C 



107. I ^S — — dx = | u 2 du, where u = sin 1 x and du 

J VI -x 2 J 



dx 



V^ 



108. J ^+J" dx = Ju 1 / 2 du, where u = tan" 1 x and du 



dx 

1 + x 2 



| u 3/2 + C = | (tan" 1 x) 3/2 + C = § ^/(tan- 1 x) 3 + C 



109 - J i5fwtt) d y = / W dy = /u du ' where u = tan_1 y 

= ln|u|+C = ln|tan- 1 y|+C 



and du = ^-j 

1+y 2 



110. / 



dy = I sin-^ ^y = I u ^ u ' wnere u = sin 1 y and du 



(sin- 1 y) Vl+y 2 

= In |u| +C = ln|sin- 1 y| +C 



dy 



in. r' sec2 ' se ;" ix) dx= r /3 s 

Ji/2 x\/x 2 -l Jir/4 



\/2 x\/x 2 - 1 Jir/ 4 

[tan u]^ 3 = tan §- tan f = ^ - 1 



sec 2 u du, where u = sec x x and du = — , dx ; x = \/2 =>• 



\/x 2 - 1 



u=f,x = 2 



1 12. / _ cos ^ — ^ dx = I cos u du, where u = sec 1 x and du = — 4 s — ; x = —, 

J 2 A/1 x\/x 2 -l Jtt/6 x\/x 2 -1 J 



u=g,x = 2 



[sinu]^ 6 = sin|-sin| = ^— 



113. lim Msziix = lim Wi-25x2; =5 

x^O x x^O ' 



1 14. lim 



n/^i - i; m M^i 



lim 



2 _ , M* 



x^l+ scu x x-*!-* 



lim 

x^ 1+ 



1) (x 2 -l)- 1 / 2 (2x) 



115. lim x tan 

X — > CO 



-1 (T\ 



lim 

X — > oo 



tan" 1 (2X" 1 ) 



116. lim 

x^0 



2 tan" 1 3x 2 



7x 2 



• r i ^ 


I 


x| v'x 2 -l / 




C -2x-2 \ 


lim 

X — > CO 


Vi+4x~ 2 y 


-X" 2 


6 


6 



lim x |x| = 1 
x — > l + 



lim . , . _ 2 

x — > CO' l+4x~ 2 



lim ,,. = lim , . 

x ^0 14x x^0 7(1+9x4) 7 



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Section 7.7 Inverse Trigonometric Functions 465 



117. Ify = lnx- | ln(l+x 2 ) - ^_p + C, then dy 



X 1 +x 2 



dx 



— I I _ x _ 1 , tan" 1 x \ 

~~ \x 1+X 2 X(l+X 2 ) + X 2 J 

which verifies the formula 



H V _ xtl+x^-x^x+ttan^xJtl+x 2 ) , _ tan" 1 x Hy 
ax- x 2(i + x2) QX - x 2 QX ' 



118. Ify=^cos-5x + |/ 7 ^dx,thendy=[x3cos-5x + (^)( 7r ^)+|( 7r ^)]dx 



(x 3 cos : 5x) dx, which verifies the formula 



119. If y = x (sin- 1 x) 2 - 2x + 2yJ \ - x 2 sin" 1 x + C, then 



j [/ ■ _i \2 2x (sin 1 x) /■, 

dy = [(sin x) + v ) i _ x2 i - 2 
the formula 



+ -^ sin" 1 x + 2^1 -x 2 ( -r±— )] dx = (sin" 1 x) 2 
V 1 — x 2 v V i — * / J 



dx, which verifies 



ln(a 2 + x 2 ) + p 2 ^ I -2 ■ : 



120. Ify = xln(a 2 +x 2 ) -2x + 2atan~ 1 (|) + C, then dy 

= [in (a 2 + x 2 ) + 2 (pipfs) -2\ dx = In (a 2 + x 2 ) dx, which verifies the formula 



1+5 



dx 



121. 



dy _ 1 

dx yr: 



dy 



dx 



yr 



=>. y = sin : x + C; x = and y = => = sin x + C => C = =>- y = sin -1 x 



dy 



122. ~g = jjL - 1 =>• dy = (y^ - l) dx =>• y = tan" 1 (x) - x + C; x = and y = 1 => 1 = tan" 1 - + C 
=>■ C = 1 => y = tan -1 (x) - x + 1 

123. ^ 



dx 

XV x 2 — 1 



dy =- ,"* =» y = sec -1 |x| + C; x = 2 and y = 7r => ir = sec -1 2 + C => C = 7r - sec -1 2 



d" x\/x 2 - 1 

= 7r-f = f => y = sec" 1 (x) + f , x > 1 



124. 



dy _ _J 2 

dx 1 + x 2 yjT 



d y=(ir^ 



v^ 



dx =>■ y = tan : x — 2 sin x x + C; x = and y = 2 



=> 2 = tan" 1 0-2 sin" 1 + C => C = 2 =>• y = tan" 1 x - 2 sin" 1 x + 2 



125. The angle a is the large angle between the wall and the right end of the blackboard minus the small angle 



126. V = tt £ 3 [2 2 - (sec y) 2 ] dy = tt [4y - tan y] q /3 = tt (% - v^) 

127. V = (i) 7rr 2 h = Q) tt(3 sin 6>) 2 (3 cos 0) = 9tt (cos - cos 3 0), where < 9 < 



S -v d_V 
2 ^ dfl 



-97r(sin(9)(l -3 cos 2 



=>• sin = or cos = ± -4- =>■ the critical points are: 0, cos l ( -4- j , and cos : ( \- j ; but 

1 I j- ] is not in the domain. When = 0, we have a minimum and when 6* = cos" 1 ( -4- j w 54.7°, we 



cos I — 

have a maximum volume 



128. 65° + (90° -P) + (90° - a) = 180° => a = 65° - (3 = 65° - tan" 1 (|) w 65° - 22.78° « 42.22° 

129. Take each square as a unit square. From the diagram we have the following: the smallest angle a has a 
tangent of 1 =>• a = tan" 1 1; the middle angle (3 has a tangent of 2 =>• /? = tan" 1 2; and the largest angle 7 
has a tangent of 3 =>• 7 = tan" 1 3. The sum of these three angles isw =4> a + (3 + j = tt 

=>- tan" 1 1 + tan" 1 2 + tan" 1 3 = tt. 

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466 Chapter 7 Transcendental Functions 

130. (a) From the symmetry of the diagram, we see that tt — sec -1 x is the vertical distance from the graph of 

y = sec -1 x to the line y = 7r and this distance is the same as the height of y = sec -1 x above the x-axis at 
— x; i.e., 7r — sec -1 x = sec -1 (— x). 
(b) cos -1 (— x) = 7r — cos -1 x, where — 1 < x < 1 =>• cos -1 (— -) = n — cos -1 (-), where x > 1 or x < —1 

=>• sec -1 (— x) = 7r — sec -1 x 

131. sin -1 (1) + cos -1 (1) = f + 0= f ; sin" 1 (0) + cos" 1 (0) = + § = | ; and sin" 1 (-1) + cos" 1 (-1) = - § + tt = f. 
If x 6 (—1,0) and x = —a, then sin -1 (x) + cos -1 (x) = sin' 1 (—a) + cos -1 (—a) = — sin -1 a + (tt — cos -1 a) 

= tt — (sin -1 a + cos -1 a) = tt — | = f from Equations (3) and (4) in the text. 



132. 




x => tan a = x and tan (3 



a + /3 



tan 1 x + tan : ; 



133. (a) Defined; there is an angle whose tangent is 2. 
(b) Not defined; there is no angle whose cosine is 2. 

134. (a) Not defined; there is no angle whose cosecant is \. 
(b) Defined; there is an angle whose cosecant is 2. 

135. (a) Not defined; there is no angle whose secant is 0. 
(b) Not defined; there is no angle whose sine is y 2. 

136. (a) Defined; there is an angle whose cotangent is — | . 
(b) Not defined; there is no angle whose cosine is —5. 

137. a(x) = cor 1 (§) - cor 1 (|) , x > =* a'(x) = ^ + ^h = ""(WH^)"^ ; solvin S 

a'(x) = => -135 - 15x 2 + 675 + 3x 2 = =4> x = 3a/5 ; a'(x) > when < x < 3^ and a'(x) < for 
x > 3 v 5 =>• there is a maximum at 3y5 ft from the front of the room 



138. From the accompanying figure, a + j3 + 6 = tt, cot a = y 
and cot /? = ^ =4> 9 = tt - cor 1 x - cor 1 (2 - x) 



1 












a 


\ B / 


'e 









X 




>. 



de _ _\ i _ i+(2-x) 2 -(i+x 2 ) 

dx 1+x 2 l+(2-x) 2 (l+x 2 )[l+(2-x) 2 ] 



= (i + x 2 )V + 4 (2-x) 2 ] solving d £=0 => x = l;g > Ofor < x < 1 and f <0forx>l 
=>• at x = 1 there is a maximum 8 = tt — cot -1 1 — cor 1 (2— 1) = tt — f — § = f 



139. Yes, sin J x and —cos : x differ by the constant | 



140. Yes, the derivatives of y = — cos x x + C and y = cos : (— x) + C are both —A 



v'l-x 2 



141. esc -1 u = f - sec -1 u => £ (csr 1 u) = £ (| - sec" 1 u) = 



j ii 






|u| v u 2 — 1 |u| v u 2 — 1 



,H>i 



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Section 7.7 Inverse Trigonometric Functions 467 



142. y = tan : x =>• tan y = x => 



£(tany)=£(x) 



(sec 2 y) 



dx 



1 



dy _ 1 
dx sec 2 y 



(^ 



1+x 2 



1+X 2 



, as indicated by the triangle 




143. f(x) = sec x =4> f'(x) = sec x tan x 



dt- 1 

dx 



dx lx = f-l(b) 



secfsec- 1 b)tan(sec-! b) b(±\/b 2 -l) ' 



Since the slope of sec x is always positive, we the right sign by writing x-sec x 



xl^x 2 -!' 



144. cot -1 u = | - tan -1 u =>• j- (cot -1 u) 



£( f _tan- 1 u)=0- T f^ 



1+u 2 



145. The functions f and g have the same derivative (for x > 0), namely r } . The functions therefore differ 

v x ' x "*" ' 



by a constant. To identify the constant we can set x equal to in the equation f(x) = g(x) + C, obtaining 
sin -1 (-1) = 2 tan" 1 (0) + C => 



|=0 + C =>■ C = - f . For x > 0, we have sin" 1 (|=i) = 2 tan" 1 \fc-\ 



146. The functions f and g have the same derivative for x > 0, namely y— ^ • The functions therefore differ by a 
constant for x > 0. To identify the constant we can set x equal to 1 in the equation f(x) = g(x) + C, obtaining 



- 1 (AS\ = tan" 1 1 + C =4> | = f + C =^> C = 0. For x > 0, we have sin" 



l l 



sfx 2 + I 



tan 



-l l 



*vs 



*vs 



147- V = tt / ^ (^i-) dx = n J_^ /3 ^ dx = tt [tan- 1 x]_^ /3 = tt [tan" 1 ^3 - tan" 1 (- f )] 
= *[f-(-!)]=T 



148. 



y = VT3rf = (l-x 8 ) 1/2 ^ y ' = (i) ( i_ x 2 ) -i/2 ( _ 2x ) => l + (y') 2 = T ^;L=/^l + (y') 2 dx 

= 2 X' /2 Tib dx = 2 I siirl ^ = 2 (1 " °) = I 



149. (a) A(x) = | (diameter) 2 = f [-^ - 

= ^[tan- 1 x]i 1 = (^)(2)(|) = f 
(b) A(x) = (edge) 2 = [-^i- - (- ^_) 



v/T^ 



1+x 2 



V=£A(x)dx=£^ 



dx^ 

+ X 2 



1+X 2 



v=/;A(x)dx=/_; 



4dx 

1+x 2 



4[tan" 1 x]i 1 = 4[tan" 1 (l)-tan" 1 (-l)] = 4 [f - (- f )] = 2tt 



150. 



(a) A(x) = g (diameter) 2 = J (^ - o)* = \ [^-^ 



\/l-x 2 



V 



£ A « 



dx 



rX 7^ dx = * [sin " x] -^ = * K 1 (^) - sin " (-#)]=- tf - (- 1)] = t 



(b) A(x ) =«i T ^ = i(^_ )- = 7 



2[sin" 1 x]! 



1 ,1 fi/ 2 



V*/2 



2(?-2}=7T 



l-x 2 



V = X A(x) dx = J 



■fin- \J\--i? 



dx 



151. (a) sec" 1 1.5 = cos" 1 ^ » 0.84107 
(c) cot" 1 2 = f - tan" 1 2 w 0.46365 

152. (a) sec" 1 (-3) = cos" 1 (- \ 



1.91063 



(c) cor 1 (-2) = f - tan" 1 (-2) rj 2.67795 



(b) esc" 1 (-1.5) = sin" 1 



(b) esc : 1.7 = sin 



1.5/ 



-0.72973 



0.62887 



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468 Chapter 7 Transcendental Functions 



153. (a) Domain: all real numbers except those having 
the form | + kn where k is an integer. 

Range: - § < y < | 



(b) Domain: — oo < x < oo; Range: — oo < y < oo 
The graph of y = tan' 1 (tan x) is periodic, the 
graph of y = tan (tan -1 x) = x for — oo < x < oo. 




y y = tan" (tan x 




3it 



■3k -k. 



-3it 



y = tan 



(tan-'x) 



k 3n 



154. (a) Domain: — oo < x < oo; Range: — | < y < | 



(b) Domain: —1 < x < 1; Range: — 1 < y < 1 
The graph of y = sin -1 (sin x) is periodic; the 
graph of y = sin (sin -1 x) = x for —1 < x < 1. 







71 

~2 


y = sin (sin x) 
/ \ 2 


-2* -22. 
2 


-it 




2. it\ /zk 
2 \S 



-2 



y = sin 



(sin- 1 x) 



1 2 



155. (a) Domain: — oo < x < oo; Range: < y < ir 



(b) Domain: —1 < x < 1; Range: — 1 < y < 1 
The graph of y = cos -1 (cos x) is periodic; the 
graph of y = cos (cos -1 x) = x for — 1 < x < 1. 



y = cos" (cos x ) 




y = cos 



\cos~ xj 



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Section 7.7 Inverse Trigonometric Functions 469 



156. Since the domain of sec -1 x is (— oo, — 1] U [1, oo), we 
have sec (sec -1 x) = x for |x| > 1. The graph of 
y = sec (sec -1 x) is the line y = x with the open 
line segment from (—1, —1) to (1, 1) removed. 



c(sec x) 



157. The graphs are identical for y = 2 sin (2 tan l x) 

= 4 [sin (tan- 1 x)] [cos (tan" 1 x)] = 4 (-^) (-^) 

^777 

= ^^j from the triangle 




y = 2 sin (2 tan -1 x) . 




158. The graphs are identical for y = cos (2 sec l x) 
= cos 2 (sec -1 x) — sin 2 (sec -1 x) = \ — ^^ 



2-x 



-j- from the triangle 




/7TT 



y 




5C 






4C 






3C 

y = cos(2sec~ 1 x) 

/ 


\ 


2-x 2 
'" x 2 


"tO =3 




— s » — 



159. The values of f increase over the interval [— 1, 1] because 
f ' > 0, and the graph of f steepens as the values of f ' 
increase towards the ends of the interval. The graph of f 
is concave down to the left of the origin where f" < 0, 
and concave up to the right of the origin where f " > 0. 
There is an inflection point at x = where f " = and 
f ' has a local minimum value. 




160. The values of f increase throughout the interval (—00, 00) 
because f ' > 0, and they increase most rapidly near the 
origin where the values off are relatively large. The 
graph of f is concave up to the left of the origin where 
f " > 0, and concave down to the right of the origin 
where f " < 0. There is an inflection point at x = 
where f " = and f has a local maximum value. 




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470 Chapter 7 Transcendental Functions 
7.8 HYPERBOLIC FUNCTIONS 



1 . sinh x : 
coth x 



| => cosh x = \f\ + sinh 2 x = J 1 + (- |) 



9_ 
16 



i = ! , tanh x = ^i 

16 4 cosh x 



tanh x 



sech x 



cosh x 



I , and csch x = -J— = — t 

5 ' sin x 3 



2. sinh x = | =>- cosh x = y 1 + sinh 2 x = -i/l 
sech x 



16 
9 



¥ = | , tanh x — 

9 3 cosh x 



sinh x (3 ) 



§ = 5> cothx =sk = i 



cosh x 



| , and csch x = ^4— = | 

5 ' sinh x 4 



3. cosh x = || , x > =>• sinh x = v cosh 2 x — 1 



289 
225 



_64 

225 ~~ 15 



*■ , tanh x = ^i - i 

' cosh x 



£ , coth x = -4— 

17 tanh x 



sech x = — \— = || , and csch x = -r\— = ~ 

cosh x 17 ' sinh x a 



m 



4. cosh x 



coth x 



13 
5 


X > 





=>- sinh x 


= v cosh 2 


*-* = Vf 


- 1 


1 


X 


13 

12 


, sech x = 


l 


5 
" 13 


, and csch x = 


l 


tanh 


cosh x 


sinh > 


vl 


- 1 


(4 


x +e -!„xN _ 


. phix , 


1 


-* + I 





144 _ 12 



sinh x 



, , tanhx — 

25 5 cosh x 



12 

13 



6. sinh (2 In x) 



e 21nx_ e -21„x 



tt2 1 I 

x "?J x 4 -l 



2x2 



7. cosh 5x + sinh 5x 



e 5x + e" 5x i e 5x - e" 5x 



Ox 



8. cosh 3x — sinh 3x = e ~ + e ' — ^- L - 



9. (sinh x + cosh x) 



4 / e x — e x I e x +e 



x\4 4x 



(e x ) 4 = e' 



10. In (cosh x + sinh x) + In (cosh x — sinh x) = In (cosh 2 x — sinh 2 x) = In 1 = 

11. (a) sinh 2x = sinh (x + x) = sinh x cosh x + cosh x sinh x = 2 sinh x cosh x 
(b) cosh 2x = cosh (x + x) = cosh x cosh x + sinh x sin x = cosh 2 x + sinh 2 x 

12. cosh 2 x - sinh 2 x = ("^f 1 ) 2 - ("^f 1 ) 2 = \ [(e x + e" x ) + (e x - e~ x )] [(e x + e" x ) - (e x - e~ x )] 

= \ (2e x ) (2e- x ) = \ (4e°) = \ (4) = 1 

13. y = 6 sinh f => & = 6 (cosh f ) (±) = 2 cosh f 



14. y = \ sinh(2x+ 1) 



%, = \ [cosh(2x + 1)](2) = cosh(2x + 1) 



15. y = 20 tanh = 21 1 / 2 tanh t 1 / 2 =>. g = [sech 2 (t 1 / 2 )] (If 1 ^) (2tV 2 ) + (tanh t 1 / 2 ) (r 1 / 2 ) 

= sech 2 0+^^ 

16. y = t 2 tanh \ = t 2 tanh t" 1 => % = [sech 2 (r 1 )] (-r 2 ) (t 2 ) + (2t) (tanh t -1 ) = - sech 2 \ + 2t tanh i 



17. y = In (sinh z) =>■ & = ^ = coth z 

•' v 7 dz sinh z 



18. y = In (cosh z) => & = ^4 = lanh z 



dz cosh z 



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Section 7.8 Hyperbolic Functions 471 



19. y = (sech 0)(1 - In sech 9) 



dy 



— sech 9 tanh i 



! (sech 0) + (- sech tanh 0)(1 - In sech 0) 
sech tanh - (sech tanh 0)(l - In sech 0) = (sech tanh 0)[1 - (1 - In sech 0)] 
(sech tanh 0)(ln sech 0) 



20. y = (csch 0)(1 - In csch 0) 



dy 



(csch 0) i 



- csch 8 coth \ 



(1 - In csch 0)(- csch coth 0) 



= csch coth - (1 - In csch 0)(csch coth 0) = (csch coth 0)(1 - 1 + In csch 0) = (csch coth 0)(ln csch 0) 

21. y = In cosh v - \ tanh 2 v => ^ = ^^ - (|) (2 tanh v) (sech 2 v) = tanh v - (tanh v) (sech 2 v) 

= (tanh v) ( 1 — sech 2 v) = (tanh v) (tanh 2 v) = tanh 3 v 

22. y = In sinh v - \ coth 2 v => ^ = f^ - (|) (2 coth v) (- csch 2 v) = coth v + (coth v) (csch 2 v) 

= (coth v) (1 + csch 2 v) = (coth v) (coth 2 v) = coth 3 v 

23. y = (x 2 + 1) sech (In x) = (x 2 + 1) (^r^) = (x 2 + 1) (^) = (x 2 + 1) (^) = 2x =► g = 2 
y = (4x 2 - 1) csch (In 2x) = (4x 2 - 1) ( e „, 2 , + 2 e _^ ) = (4x 2 - 1) (j^h^) = ^ ~ ! ) (sb) 



24 



4x =► & = 4 

dx 



25. y = sinh- 1 ^ = sinh" 1 (x 1 / 2 ) => £ = 2 ^ = ^ = -U ■ 

J V V / dx V 1+(x i /2) 2 2V^\/TT^ 2 x /x(l+x) 



26. y = cosh" 1 2a/x + 1 = cosh" 1 (2(x + l) 1 



/ 2 1 



dy= (2)(l)(x + l)-^ = 1 = l 

dx vW+D 1/2 ] 2 -l V^+lV^+3 \/4x2 + 7x + 3 



27. y = (1 -0)tanh" 1 



-1 a _*. & - 



(1 - ^) (i3p) + (-1) tanh_1 * = TTe ~ tanh_1 * 



28. y = (0 2 + 20) tanh" 1 (0+1) => | = (0 2 + 20) [ 1 _ (9 1 +1)2 ] + (20 + 2) tanh" 1 (0 + 1) 
- -^±2»_ _l (261 + 2) tanh" 1 (0 + 1) = (20 + 2) tanh" 1 (0 + 1) - 1 



-29 



29. y = (1 - ^coth^v/t = (1 - Ocottr 1 (t 1 / 2 ) 



i=d-t) 



(i)'- i/2 

1 - (tl/2) 2 



+ (-l)coth- 1 t 



1 ft 1 / 21 ! 



1 

2\/t 



coth _1 \/t 



30. y = (1 -t 2 )coth" 1 t => f = (1 -t 2 ) (y^) + (-2t) coth" 1 1 = 1 -2tcoth- 1 t 



31. y = cos x x — x sech x x => 
= — sech -1 x 



dy _ -1 
dx Vl-x 2 



(—7==) + (1) sech" 1 xj = -j^= + -4— - sech" 1 x 



32. y = In x + a/1 — x 2 sech : x = In x + (1 — x 2 ) sech 1 x => — 
= i + C 1 " x2 ) 1/2 (xTlb) + (J) C 1 " ^) _1/2 (-2x) sech- 1 x 



dx 
-1 „ _ 1 1 



y/T- 



sech x 



v/T" 



sech x 



33. y = csch" 1 (±) 



dy 



In (1)- In (2) _ In 2 



i+4 



i+4 



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472 Chapter 7 Transcendental Functions 

,-1 off -^ d Y _ On 2)2° _ -In 2 



34. y = csch" 1 2" => 



1 ^l + (2») 2 V^h - : 



35. y = sinh x (tan x) => 



dy 



sec 2 x _ sec 2 x _ sec x| sec x 



sec X 



dx ^1 +(tanx) 2 y- sec^ x 

36. y = cosh- 1 (sec x) =>■ £ = ( 7 x) , (tanx) = (sec , x)( ' an x) = (se " x)(ta " x) = sec x, 0< x < f 

J V ' dx \/sec 2 x - 1 \Aan 2 x I 1 ™ x l 2 

37. (a) If y = tan" 1 (sinh x) + C, then f = , :° sh x = ^* = sec h x, which verifies the formula 

v ' J v 7 ' dx 1 + sinn^ x cosh z x 



(b) If y = sin -1 (tanh x) + C, then 



djr _ sech 2 x — _ sech^x _ secll w iji cn verifies the formula 

dx V 1 - tanh 2 x secn x 



38. If y = f sech" 1 x - \ y/l - x 2 + C, then g = x sech" 1 x + f 



2 J ^" * 2 

which verifies the formula 



2 f -1 1 | 2x 

2 Vxv'l-x 2 / 4\/l -x 2 



x sech : x, 



39. If y = ^=1 com- 1 x + | + C, then -| = x coth -1 x + ( ^y 1 J (73^2) + 5 = * coth -1 x, which verifies 
the formula 

40. If y = x tann -1 x + \ In (1 - x 2 ) + C, then -| = tanh -1 x + x (7^2) + 5 (7^2) = tanh -1 x, which verifies 
the formula 

41. I sinh 2x dx = \ J sinh u du, where u = 2x and du = 2 dx 



cosh u 



+ c 



cosh 2x 



+ c 



42. I sinh I dx = 5 / sinh u du, where u = | and du = i dx 



5 cosh u + C = 5 cosh | + C 



43. I 6 cosh (| — In 3) dx = 12 I cosh u du, where u = | — In 3 and du = | dx 
= 12 sinh u + C = 12 sinh (§ - In 3) + C 



44. / 4 cosh (3x — In 2) dx = | I cosh u du, where u = 3x — In 2 and du = 3 



dx 



| sinh u + C = 5 sinh (3x - In 2) + C 



45. J tanh f dx = 7 J ^*jl du, where u = f and du = i dx 
= 7 In |cosh u| + Ci = 7 In |cosh || + Ci = 7 In 

= 7 1n|e x / 7 +e- x / 7 |+C 



/'+eW 



Ci = 7 In |e x / 7 + e- x / 7 | - 7 In 2 + Ci 



46. f coth 4? d6> = \/3 f ^ du, where u = 4? and du 

J ,/3 V J smhu ' ,/3 



V^ 



75 



e«/V3- e -9/\/3 I 



+ Ci 



a/3 In |sinhu| + Ci = a/3 In sinh 4- + Ci = y/3 In , 

a/3 In |e 9 /v^ - e-^/v^l - a/3 In 2 + d = a/3 In le^/v^ - e- fl /^| + C 



47. I sech 2 (x — |) dx = I sech 2 u du, where u = (x — |) and du = dx 
= tanh u + C = tanh (x - i) + C 



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Section 7.8 Hyperbolic Functions 473 



48. I csch 2 (5 — x) dx = — / csch 2 u du, where u = (5 — x) and du = — dx 
= -(- coth u) + C = coth u + C = coth (5 - x) + C 



49. / 



sech y t tanh yt 



— dt = 2 I sech u tanh u du, where u = \J\ = t 1 / 2 and du 
: 2(- sech u) + C = -2 sech y^t + C 



dt 
2^1 



50. / cschdnt^cothdnt) ^ = J ^ Q CQth u ^ where u = h t and dll = f 

= — csch u + C = — csch (In t) + C 

51. f° 4 cothxdx= r'' ,4 £2^dx= C*'* ± du = [In \u\}l 5 /, s = In \%\ - In ||| = In \% • \ I = In f , 

Jl„2 J In 2 sinhx J 3/4 u L I I J 3/4 181 141 18 31 2 ' 

where u = sinh x, du = cosh x dx, the lower limit is sinh (In 2) = e " ~ e = — ~- = | and the upper 

.. .. A-(k\ 

limit is sinh (In 4) 



52. J o " tanh 2x dx = // ^ dx = \ J] I du = i [In |u|] 1 7/8 = ± [in (f ) - In l] = I In f , where 

u = cosh 2x, du = 2 sinh (2x) dx, the lower limit is cosh 0=1 and the upper limit is cosh (2 In 2) = cosh (In 4) 



-In 4 I p -ln4 



53. r ,n2 2e coshede= r >n2 2t e ( e ^f^)dd= r' nl {e 2S +\)d9=\^+e} '" 2 

J-ln4 J-ln4 \ 2 J J-ln4 v ' L 2 J-ln4 

= (^ - In 2) - (^ - In 4^ = (| - In 2) - (^ - In 4) = ^ - In 2 + 2 In 2 = ^ + In 2 

54. / o ,n2 4e-" sinh (9 d6» = J^e"" f^ 5 ^) d0 = 2 ^'"'(l - er 29 ) d9 = 2 \o + ^1 " 

= 2[(ln2+£=^) - (0+£)] =2(ln2+i-i) = 21n2+±-l=ln4-| 

55. /"^cosh (tan 0) sec 2 (9 d(9 = J_',cosh u du = [sinh u] \ = sinh(l) - sinh(-l) = ( s ^ £=i ) - (^^f^) 



e — e x — e * +e 



e — e , where u = tan 9, du = sec 2 9 d9, the lower limit is tan I 



1 and the upper 



limit is tan (f J = 1 



56. I 2 sinh (sin 9) cos 9 d9 = 2 I sinh u du = 2 [cosh u] = 2(cosh 1 — cosh 0) = 2 ( - 



■ - 1 
e + e _1 — 2, where u = sin 9, du = cos 9 d6, the lower limit is sin = and the upper limit is sin (f ) = 1 



57. J' c -^f^l dt = / o '" 2 cosh u du = [sinh u]^ 12 = sinh (In 2) - sinh(0) = e ' n2 - e ~'" 2 - 
u = In t, du = j dt, the lower limit is In 1 = and the upper limit is In 2 

58. f * *"")/* dx = 16j^ cosh u du = 16 [sinh u] \ = 16(sinh 2 - sinh 1) = 16 

= 8 (e 2 - e- 2 - 
limit is v 4 = 2 



— i -1 

1 y i = j , where 



(■=**)] 



(e 2 — e 2 — e + e x ) , where u = yfx = x 1 / 2 , du = | x 1//2 dx = 5=7- , the lower limit is \J 1 = 1 and the upper 



59. /cosh 2 (| ) dx = J_" ln2 cosh 2 x + 1 dx = 1 J_° (cosh x + 1) dx = \ [sinh x + x] ^ ln2 

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474 Chapter 7 Transcendental Functions 



\ [(sinh + 0) - (sinh(- In 2) - In 2)] = \ [(0 + 0) - ( e "' n2 ~ e ' n2 - In 2)] = \ 



-2 



In 2 



I(l-I+ln2) 



3 + I ln2= | +lnv ^ 



60. / o ' n '°4 sinh 2 (I) dx = J o '°'°4 ( cosh 2 x ~' ) dx = 2j o ln10 (cosh x - 1) dx = 2 [sinh x- x]J 

= 2[(sinh(ln 10) - In 10) - (sinh - 0)] = e lnl ° - e-" ,lu - 2 In 10 = 10 - ^ - 2 In 10 = 9.9 - 2 In 10 



1 In 10 

Jo 



61. sinh 



63. tanh 



-1 /-5a 



In 



+ 



25 
144 



=m( 



62. cosh" 1 (f 



(f)=m(! + yf r T) 



In 3 



-i(_n = 1 in ( ±M) 



In 3 

2 



64. coth- 1 (|) 



Iln(^)=Iln9 = ln3 



65. sech 



1( l)=ln(i±Vg^) 



In 3 



66. csdi" 1 (- -±\ = In ( - y^ ■ 



4/3 



/x/3) 



In 



(-V3 + 2) 



67. 



< a > X 



2^ 



y^ 



[sinh" 1 |] 'f = sinh" 1 a/3 - sinh = sinh" 1 a/3 



(b) sinh" 1 a/3 = In (a/3 + a/3 + l) = In (a/3 + 2\ 



68. (a) / 



1/3 



6 dx 



2 / / t x o , where u = 3x, du = 3 dx, a = 1 

Jo ^a 2 + u 2 



\/l +9x 2 Jo y'a 

= [2 sinh- 1 u] = 2 (sinh" 1 1 - sinh" 1 0) = 2 sinh" 1 1 
(b) 2 sinh" 1 1 = 2 In (l + a/I 2 + l) = 2 In (l + a/2~) 



-1 5 
4 



69. (a) J 4 y^U, dx = [coth- 1 x] L = coth- 1 2 - coth 
(b) coth" 1 2 - coth" 1 I = i [in 3 - In (fg)] = i In i 



70. (a) J ~ r ^ 3 dx = [tanh- 1 x] / 2 = tanh" 1 \ - tanh" 1 = tanh" 1 \ 

l + (l/2) \ _ 1 
1 — (1/2) ^ 2 



(b) tanh- 1 i = iln(|^^;'l = i In 3 



71. 



< a > /, 



3/13 



dx 



X 



12/13 



1/5 x\/l - 16x 2 J 4/5 u\/a 2 -u : 

[- sech" 1 ul/'f = - sech- 1 % + sech" 1 f 



, du „ , where u = 4x, du = 4 dx, a = 1 



(b) - sech" 1 % + sech" 1 \ = - In 



/ l + x/l- (12/13) 2 \ , / l + x/l-(4/5)2 \ 
^ (12/13) J 1" m \^ (4/5) ^ 



In 



( 13 + ^f IT¥ )+ln( 



5 + x/25- 16 
4 



In (5±3) _ In (13±5) = In 2 -In I 



In (2 • § ) = In \ 



72. 



« r^fe=t^ csch 



- L !| |] \ = - i (csch- 1 1 - csch" 1 \) = \ (csch- 1 | - csch" 1 l) 



(b) I (csch- 1 I - csch- 1 1) = I [in (2 + g) - In (l + a/2)] = I In (f±$) 

, cos x , dx = I , du = [sinh -1 ul n = sinh -1 — sinh -1 = 0, where u = sin x, du = cos x dx 

v/l+sin 2 x Jo v/l+u 2 1 JO 

(b) sinh" 1 - sinh" 1 = In (o + \/0 + l) - In (o + a/0 + l) = 

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Section 7.8 Hyperbolic Functions 475 



74. (a) /' 



dx 



1 x\/l+(lnx) 2 ~ Jo / 
1 




I , ** u ^ , where u = In x, du = - dx, a = 1 

Jo va 2 + u 2 x 



sinh 1 u] = sinh : 1 — sinh : = sinh * 1 



(b) sinh" 1 1 - sinh" 1 = In (l + a/1 2 + l) - In (o + \/0 2 + l) = In (l + y/l\ 



75. (a) Let E(x) = f(x)+ / ( - x) and O(x) = **>-*-*> . Then E(x) + O(x) = f(x) + f( ~ x) + f(x) ~ 2 f( ~ x) 

._ 2fW _ f(x) AlsQj E( _ x) = f(-x) + f(-(-x)) = f(x) + f(-x) = E(x) ^ E(x) j s even ^ and 



2 * vv . *„„v-, ^y ~j 2 2 

f(-x) - f(-(-x)) _ _ f(x) - f(-x) 

2 2 



O(-x) = n x > ^ * x " = - IW 2 *> = - O(x) => O(x) is odd. Consequently, f(x) can be written as 
a sum of an even and an odd function. 



(b) f(x) 



f(x) + f(-x) 



because 



f(x)-f(-x) 



2 ^v~««ov 2 

2f(x) 



if f is even and f(x) 



f(x)-f(-x) 



because 



f(x) + f(-x) 



iff is odd. 



Thus, if f is even f(x) = ^ + and if f is odd, f(x) = + ^ 



76. y = sinh l x =>- x = sinh y =>- x 



=> 2x = e y - ^ =>• 2xe y = e 2y - 1 => e 2y - 2xe y -1=0 



2x±y / 4x 2 + 4 



=>• e y = x + v x 2 + 1 =>• sinh 1 x = y = ln(x+ y x 2 + 1 ) • Since e y > 0, we cannot 



choose e y = x — y x 2 + 1 because x — y x 2 + 1 < 0. 



77. (a) v = A /^tanhf ,/* t ) =* 



dv _ /mg 
dt ~~ V k 



sech - 



2 ./|k 



2 ./|k 



t • 



Thus m^ = mg sech 2 ( y ^- t J = mg ( 1 — tanh 2 
when t = 0. 



= g sech - 
mg — kv 2 . Also, since tanh x = when x = 0, v = 



(b) lim v= lim ,/2I tanh ( J^t 
' t->oo t->oo v k ' v ' m 



^ lim tanh 

K t — > oo 



mg /n / mg 

k U; — A/ I, 



( C ) V 0555 ~ V 



i=f°° = m = 80 ^ w 178 89 ft/sec 



78. (a) s(t) = a cos kt + b sin kt 



ds 



-ak sin kt + bk cos kt =4> ^f = — ak 2 cos kt — bk 2 sin kt 



-k 2 (a cos kt + b sin kt) = — k 2 s(t) =4> acceleration is proportional to s. The negative constant — k 2 



implies that the acceleration is directed toward the origin. 



(b) s(t) = a cosh kt + b sinh kt 



ds 



ak sinh kt + bk cosh kt 



|f = ak 2 cosh kt + bk 2 sinh kt 



= k 2 (a cosh kt + b sinh kt) = k 2 s(t) =S> acceleration is proportional to s. The positive constant k 2 implies 
that the acceleration is directed away from the origin. 



79. 



=>- y = I i l - dx + / , x ■ dx => y = sech 1 (x) - yl - x 2 + C; x = 1 and 

J x\/1 — x 2 J */ 1 _ x 2 



Q- = -1 _|_ X _ _ _^^ = u x - -^ 

d " x\/l-x 2 \/l-x 2 " ^ ^ x\/l-x 2 "^ """ J \/l 

y = 0^C = 0=>y = sech" 1 (x) - v 7 ! - x 2 



nb . 

80. To find the length of the curve: y = i cosh ax =>• y' = sinh ax => L = / y 1 + (sinh ax) 2 dx 

^*b . pb 

=>■ L = I cosh ax dx = [- sinh axl = - sinh ab. Then the area under the curve is A = | - cosh ax dx 

Jo LaJoa Ji a 

= [4 sinh ax] = 4 sinh ab = Q) Q sinh ab) which is the area of the rectangle of height 1 and length L 
as claimed. 

81. V = 7T J (cosh 2 x - sinh 2 x) dx = tt J 1 dx = 2tt 



82. V 



f 1 

2tt 
Jo 



lnv/3 



sech x dx = 2n ftanh x 



ln\/3 



27T 



/3- 1A/3 



'3 + 1/V3 



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476 Chapter 7 Transcendental Functions 

pin \/5 pin \/5 

83. y = \ cosh 2x => y' = sinh 2x=^L=| yT+ (sinh 2x) 2 dx = I cosh 2x dx = [| sinh 2x] 

ln\/5 



lnV5 




1(^)] -JP-J) 



84. (a) Let the point located at (cosh u, 0) be called T. Then A(u) = area of the triangle AOTP minus the area 

I pcoshu I 

under the curve y = \J x 2 — 1 from A to T => A(u) = \ cosh u sinh u I \J x 2 — 1 dx. 
(b) A(u) = \ cosh u sinh u J V x 2 — 1 dx =>■ A'(u) = | (cosh 2 u + sinh 2 u) — ( y cosh 2 u 1 J (sinh u) 



| cosh 2 u + ^ sinh 2 u — sinh 2 u = \ (cosh 2 u — sinh 2 u) 



(1) 



(c) A'(u) = \ => A(u) = I + C, and from part (a) we have A(0) = =>• C = =>• A(u) = § => u = 2A 



85. y = 4 cosh f => 1 + (^V = 1 + sinh 2 (|) = cosh 2 (|) ; the surface area is S = J^" 27ry W 1 + f^V dx 
= 8tt P ' cosh 2 (f ) dx = 4tt f ' (l + cosh § ) dx = 4tt [x + 2 sinh §1 '° 81 

J-I11I6 \4/ J-lnI6^ }> L 2 J —In 16 

= 47r[(ln81 +2 sinh (^-)) - (-In 16 + 2 sinh (^f^))] =47r[ln(81 • 16) + 2 sinh (In 9) + 2 sinh (In 4)] 
= 4tt [In (9 - 4) 2 + (e ln9 - e" 1 " 9 ) + (e'» 4 - e" 1 " 4 )] = 4tt [2 In 36 + (9 - \) + (4 - \)] = 4tt (4 In 6 + f + f ) 



: 4tt (4 In 6 + 32Q 3 + 135 ) = 16tt In 6 + ^ 



86. (a) y = cosh x => ds = ^/(dx) 2 + (dy) 2 = ^/(dx) 2 + (sinh 2 x) (dx) 2 = cosh x dx; M x = J ^ y ds 

= J^cosh x ds = J '* 2 cosh 2 x dx = /'"'(cosh 2x + 1) dx = [^^ + x] j" 2 == | (e ln4 - e" 1 " 4 ) + In 2 



16 



In 



r ln2 . r ln2 

2;M = 2J VI +sinh 2 xdx = 2 J cosh x dx = 2 [sinh x]^' 1 " = e 1 " 2 



1 3 



Therefore, y 
(b) x = 0,y « 1.09 



M* _ ({f+'" 2 ) _ 5 , In 



= I + 1j y > ar, d by symmetry x = 0. 



(-In 2, 1.25)' 



,,(0,1.09} 



0,8 
0.6 
0.4 
0.2 



~3tr 



(ln2, 1.25) 



y = cosh x 



) 02 6!4 0'6 — x 



87. (a) y = S cosh (| x) =* tan = g = (S) [g sinh (g x)] = sinh (g x) 

(b) The tension at P is given by T cos </> = H =>■ T = H sec </> = H^/l + tan 2 . 
= H cosh (gx)=w(S) cosh (gx)=wy 



H v / l + (sinhgx)^ 



s = - sinh ax => sinh ax = as => ax = sinh : as =>■ x = - sinh J as; y = 1 cosh ax = 7 y cosh 2 
= 1 Vsinh 2 ax + 1 = 1 v^VTT = A /s 2 + A 



ax 



(a) Since the cable is 32 ft long, s = 16 and x = 15. From Exercise 88, x = - sinh l as => 15a = sinh : 16a 
=4> sinh 15a = 16a. 



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Chapter 7 Practice Exercises 477 



(b) The intersection is near (0.042, 0.672). 



y - 16a 




0.61*3 



(c) Newton's method indicates that at a rj 0.0417525 the curves y = 16a and y = sinh 15a intersect. 

(d) T = wy w (2 lb) (0^7325) » 47.90 lb 

(e) The sag is icosh(15a) - \ « 4.85 ft. 32 ' 

30 

~28 

26 



22 
20 



-£5 =10 : 3~ 




y = (23.951 )cosh(0.4175 x) 

-5 — is — 1$ 



CHAPTER 7 PRACTICE EXERCISES 

1. y = lOe"*/ 5 => I = (10) (- |) e""/ 5 = -2e-*/ 5 

2. y = v/2e^ => | = (y/l) (y/l) e^ x = 2e^ 



3 1 

■ y = i xe 



1 „„*t 1 „4x __, dy 1 r / 1 M . a 4x/i\l 1 //|_43t\ „~4x _j_ 1 „4x 1 „4x „„4x 



[x (4e 4x ) + e 4x (l)] - i (4e 4x ) = xe 4x + ± e 4 



4. y = x 2 e" 2 / x = xV 2x_1 => g = x 2 [(2x~ 2 ) e" 2x ~'] + e" 2x ~'(2x) = (2 + 2x)e" 2x ~' = 2e" 2 / x (l + x) 

5. y = In (sin 2 9) => | = 2(sin g) 2 (c fl os e) = ^ = 2 cot 

-^ V / dff sin 2 sin 8 



6. y = In (sec 2 



dy 2(sec 0)(sec tan ( 



2 ten 6 



7. y = log 



dy j_ _j^ 1 2 

dx In 2 I (*2\ I (ln2)x 



y = log 5 (3x - 7) 



In 5 



dx \\n.5> V3x-7 7 (In 5)(3x-7) 



9. y 



^ = 8- 1 (ln8)(-l) = -8-'(ln8) 



J2i . dy _ Q2t 



10. y = 9 21 => f = 9 2, (ln 9)(2) = 9 2, (2 In 9) 



11. y = 5x 3 - 6 => | = 5(3.6)x 2 - 6 = 18x 2 - e 



12. y = v /2x-^ => | = (v^)(->/2 



cf-v^-O = -2xM-0 



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478 Chapter 7 Transcendental Functions 

13. y = (x + 2)*+ 2 =>■ lny = ln(x + 2)*+ 2 = (x + 2)ln(x + 2) => £ = (x + 2) (^) + (l)ln(x + 2) 

=* |=(x + 2)*+ 2 [ln(x + 2)+l] 

14. y = 2(lnx)*/ 2 => In y = In [2(ln x)*/ 2 ] = In (2) + (|) In (In x) =► J = + (§) [£*] + (In (In x)) (±) 

=* y' = [21b + (I) ln ( ln x )] 2 ( ln x ) x/2 = ( ln x ) x/2 t ln ( ln x > + £] 



15. y = sin : \f\ - u 2 = sin l (1 - u 2 ) 1 ' 2 =^ 



l n _,^/2 _^ dy _ i(l-uT' /2 (-2u) 
du ' 



l-[(l-u 2 ) 1/2 ] 



==,0< u< 1 



uyl — u 2 v 1 — u : 

16. y = sin" 1 (-U = sin" 1 v" 1 / 2 =* | 



_|v"3/2 



2 Vl-u 2 \/l-(l-u 2 ) lulVl-u 2 



-^ 



./l _ (v-i/2) 2 2vV 2 \/l-v-i 2v 3/2 Jl^i 2v 3 / 2 v/^T 



2vVv - 1 



17. y = ln(cos : x) => y' 



:os x V 1 — x 2 COS -1 X 

18. y = z cos -1 z - Vt - z 2 = z cos -1 z - (1 - z 2 ) 1 ' 2 =4> ^ 



: — COS Z ^ 

dz yi _ z 2 



(I)(l-z 2 )- 1/2 (-2z) 



7T- 



'l-z 2 



19. y = t tan" 1 1 - (|) ln t ^ % = tan" 1 1 + t (^ 

20. y = (1 + t 2 ) cor 1 2t => I = 2t cor 1 2t+ (1 + t 2 ) (y^ 



P )-a)(^)=tan- 1 t +T ^-l 



.-1,_ C,2 _ nV2 . dy 



21. y = z sec : z — y z 2 — 1 = z sec : z — (z 2 — l) 1/z = 
+ sec -1 z 



z 2 - 1 Vz 2 - 1 



dz 

1 ' ^- - 1 - sec -1 z, z > 1 



; (n^r^) + ( sec_1 z ) (D - \ ( z2 - i) _1/2 (2z) 

\ |z| v z — 1 / 



Vz 2 -1 



22. y = 2a/x- 1 sec" 1 0c = 2(x - l) 1 



/ 2 sec- 1 fx 1 / 



dy _ 9 
dx — ^ 



(!) (x - 1)-V 2 sec- 1 (x 1 / 2 ) + (x - l) 1 / 2 [ -Qfc^ 



xy x — 1 



(sec * 
27T 



1 ■ 2x 7 x/x-i x 



2 \ sec 



23. y = esc 1 (sec 0) 



dy 



— sec 8 tan _ tan fl __ i C\ ^ f) ^ H 

|sec 0| ./sec 2 9 - 1 " l tan "I ~ ' 2 



24. y = (1 + x 2 ) e tan x => y' = 2xe' 



_lx + (l + x 2 )(^5) =2xe tan " lx + 



tan L x I ~tan L x 



25. y = ^^lny = ln(^)=ln(2) + ln(x 2 + l)-Iln(cos2x)^^=0+^ r -(i) 



'1^ (-2 sin 2x) 
cos 2x 



/'=(^T 



+ tan 2x) y 



2(x 2 + l) I 2x 
Vcos 2x Vx 2 + 1 



tan 2x) 



26 - y = yir^ => ln y = ln T^ = h M 3x + 4 ) - ln ( 2x - 4 )1 



y^ - J_ ( 3 



y 10 V3x + 4 2x-4V 



v ' _ J_ f_J L_\ v _ io/ 3x + 4 / 1 \ /_3 1.^ 

* 10 V3x + 4 x-2) J V 2x-4 VW V3x + 4 x-2-* 



Cfigl (c) 1 Pearson Education, Inc., publishing as Pearson Addison-Wesle 



Chapter 7 Practice Exercises 479 



27. y= [ fr|§^j ] 5 => hi y = 5 [ln(t + 1) + ln(t- 1) - ln(t - 2) - ln(t+ 3)] 



9* 



^ Vt+1 ~ t-1 t-2 t + 3/ 



dy g (t+l)(t-l) M, J l i_\ 

dt J I (t-2)(t + 3) I Vt+l T t-l t-2 t + 3/ 1 



28 ' y = #h =* Iny = ln2 + lnu + uln2-lln(u 2 + l) => (l) (|) = I +ln2 - \ (^ 



\/u 2 +l 

dy = 

du x / u2 + 1 



--^^^-^ 



29. y = (sin0)^ =► In y = y/e In (sin 6) => (i) (|) = v^ (^|) + ± 0" 1 / 2 in (sin 0) 



=> | = (sin 0)V" ( V^ cot 9 + 



In (sin 9) 



i /n 



30. y = (lnx) 1 /inx ^ l n y = (^) ln(ln x) => £ = (£) (£) (I) + In (In x) L(]nx)2J uy 

^y' = (lnx)^x[i_i^] 

31. J e x sin (e x ) dx = J sin u du, where u = e x and du = e x dx 

= — cos u + C = — cos (e x ) + C 

32. J e' cos (3e l — 2) dt = I J cos u du, where u = 3e l — 2 and du = 3e l dt 

= ± sin u + C = i sin (3e l - 2) + C 

33. J e x sec 2 (e x — 7) dx = J sec 2 u du, where u = e x — 7 and du = e x dx 

= tan u + C = tan (e x - 7) + C 

34. J e y esc (e y + 1) cot (e y + 1) dy = J esc u cot u du, where u = e y + 1 and du = e y dy 

= - esc u + C = - esc (e y + 1) + C 

35. J (sec 2 x) e tanx dx = J e u du, where u = tan x and du = sec 2 x dx 

= e u + C = e tanx + C 

36. J (esc 2 x) e cotx dx = — J e u du, where u = cot x and du = — esc 2 x dx 

= -e u + C = -e cotx + C 



37. I ., : . dx = \ I - du, where u = 3x — 4, du = 3 dx; x = — 1 =>• u = —7, x = 1 =>■ u 

J_! 3x-4 3J_ 7 u 

= | [In |u|] Z\ = | [In |-1| - 1b |-7|] = I [0 - In 7] = - f 

38. f ^P